ML Aggarwal Solution Class 10 Chapter 7 Ratio and Proportion Test

 Test

Question 1

Find the compound ratio of:

$(a+b)^{2}:(a-b)^{2}$

$\left(a^{2}-b^{2}\right):\left(a^{2}+b^{2}\right)$

$\left(a^{4}-b^{4}\right):(a+b)^{4}$

Sol :

$(a+b)^{2}:(a-b)^{2}$

$\left(a^{2}-b^{2}\right):\left(a^{2}+b^{2}\right)$

$\left(a^{4}-b^{4}\right):(a+b)^{4}$

$=\frac{(a+b)^{2}}{(a-b)^{2}} \times \frac{a^{2}-b^{2}}{a^{2}+b^{2}} \times \frac{a^{4}-b^{4}}{(a+b)^{4}}$

$=\frac{(a+b)^{2}}{(a-b)^{2}} \times \frac{(a+b)(a-b)}{a^{2}+b^{2}}\times \frac{\left(a^{2}+b^{2}\right)(a+b)(a-b)}{(a+b)^{4}}$

$=\frac{1}{1}$ = 1 : 1

Question 2

If (7 p + 3 q) : (3 p – 2 q) = 43 : 2 find p : q

Sol :

(7p + 3q) : (3p – 2q) = 43 : 2

$\Rightarrow \frac{7 p+3 q}{3 p-2 q}=\frac{43}{2}$

$\Rightarrow \quad 129 p-86 q=14 p+6 q$

$\Rightarrow \quad 129 p-14 p=6 q+86 q$

$\Rightarrow \quad 115 p=92 q$

$\Rightarrow \quad \frac{p}{q}=\frac{92}{115}=\frac{4}{5}$

∴p : q= 4 : 5 

Question 3

If a : b = 3 : 5, find (3a + 5b): (7a – 2b).

Sol :

a : b = 3 : 5

$\Rightarrow \frac{a}{b}=\frac{3}{5}$

⇒ 3a + 5n : 7a – 2b

Dividing each term by b

$3 \frac{a}{b}+5: 3 \frac{a}{b}-2$

$\Rightarrow 3 \times \frac{3}{5}+5: 7 \times \frac{3}{5}-2$

$\Rightarrow\left(\frac{9}{5}+5\right):\left(\frac{21}{5}-2\right)$

$\Rightarrow \frac{9+25}{5}: \frac{21-10}{5}$

$\Rightarrow \frac{34}{5}: \frac{11}{5}$

=34:11

Question 4

The ratio of the shorter sides of a right angled triangle is 5 : 12. If the perimeter of the triangle is 360 cm, find the length of the longest side.

Sol :

Let the two shorter sides of a right-angled triangle be 5x and 12x.

Third (longest side)

$=\sqrt{(5 x)^{2}+(12 x)^{2}}$

$=\sqrt{25 x^{2}+144 x^{2}}$

$=\sqrt{169 x^{2}}=13 x$

But 5x+12x+13x=360cm

⇒30x=360

$\Rightarrow x=\frac{360}{30}=12$

∴Length of the longest side=13x

=13×12cm=156cm

Question 5

The ratio of the pocket money saved by Lokesh and his sister is 5 : 6. If the sister saves Rs 30 more, how much more the brother should save in order to keep the ratio of their savings unchanged?

Sol :

Let the savings of Lokesh and his sister are 5x and 6x.

and the Lokesh should save Rs y more Now, according to the problem,

$\Rightarrow \frac{5 x+y}{6 x+30}=\frac{5}{6}$

⇒ 30x + 6y = 30x + 150

⇒ 6y=150

$\therefore y=\frac{150}{6}=25$

Hence, Lokesh should save 25 more

Question 6

In an examination, the number of those who passed and the number of those who failed were in the ratio of 3 : 1. Had 8 more appeared, and 6 less passed, the ratio of passed to failures would have been 2 : 1. Find the number of candidates who appeared.

Sol :

Let number of passed = 3 x

and failed = x

Total candidates appeared = 3x + x = 4x.

In second case

No. of candidates appeared = 4 x + 8

and No. of passed = 3 x – 6

and failed = 4x + 8 – 3x + 6 = x + 14

then ratio will be = 2 : 1

Now according to the condition

$\frac{3 x-6}{x+14}=\frac{2}{1}$

⇒ 3x-6=2x+28

⇒ 3x-2x=28+6

⇒ x=34

∴No. of candidates appeared

=4x=4×34=136

Question 7

What number must be added to each of the numbers 15, 17, 34 and 38 to make them proportional ?

Sol :

Let x be added to each number, then numbers will be

15 + x, 17 + x, 34 + x, and 38 + x.

Now according to the condition

$\frac{15+x}{17+x}=\frac{34+x}{38+x}$

$\Rightarrow(15+x)(38+x)=(34+x)(17+x)$

$\Rightarrow 570+53 x+x^{2}=578+51 x+x^{2}$

$\Rightarrow x^{2}+53 x-x^{2}-51 x=578-570$

$\Rightarrow 2 x=8 \Rightarrow x=4$

$\therefore 4$ is to be added.

Question 8

If (a + 2 b + c), (a – c) and (a – 2 b + c) are in continued proportion, prove that b is the mean proportional between a and c.

Sol :

(a + 2 b + c), (a – c) and (a – 2 b + c) are in continued proportion

$\Rightarrow \frac{a+2 b+c}{a-c}=\frac{a-c}{a-2 b+c}$

$\therefore \frac{a+2 b+c}{a-c}=\frac{a-c}{a-2 b+c}$

$\Rightarrow \quad(a+2 b+c)(a-2 b+c)=(a-c)^{2}$

$\Rightarrow \quad a^{2}-2 a b+a c+2 a b-4 b^{2}+2 b c+a c-2 b c+c^{2}=a^{2}-2 a c+c^{2}$

$\Rightarrow \quad a^{2}-2 a b+a c+2 a b-4 b^{2}+2 b c+a c-2 b c+c^{2}-a^{2}+2 a c-c^{2}=0$

$\Rightarrow \quad 4 a c-4 b^{2}=0 \Rightarrow a c-b^{2}=0$

$\Rightarrow \quad b^{2}=a c$

Here b is the mean proportional between a and c

Question 9

If 2, 6, p, 54 and q are in continued proportion, find the values of p and q.

Sol :

2, 6, p, 54 and q are in continued proportional then

$\Rightarrow \frac{2}{6}=\frac{6}{p}=\frac{p}{54}=\frac{54}{9}$

(i) $\because \frac{2}{6}=\frac{6}{p}$ then 

2p=36

$\Rightarrow p=18$

(ii) $\frac{p}{54}=\frac{54}{q} \Rightarrow p q=54 \times 54$

$\Rightarrow 18 q=54 \times 54$

$\Rightarrow q=\frac{54 \times 54}{18}=162$

Hence , p=18, q=162

Question 10

If a, b, c, d, e are in continued proportion, prove that: $a: e=a^{4}: b^{4}$

Sol :

a, b, c, d, e are in continued proportion

$\Rightarrow \frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}=k($ say $)$

$d=e k, c=e k^{2}, b=e k^{3}$ and $a=e k^{4}$

Now, $\mathrm{L.H.S.}=\frac{a}{e}=\frac{e k^{4}}{e}=k^{4}$

R.H.S $\frac{a^{4}}{b^{4}}=\frac{\left(e k^{4}\right)^{4}}{\left(e k^{3}\right)^{4}}=\frac{e^{4} k^{16}}{e^{4} k^{12}}$

$=k^{16-12}=k^{4}$

∴L.H.S=R.H.S

Question 11

Find two numbers whose mean proportional is 16 and the third proportional is 128.

So :

Let x and y be two numbers

Their mean proportion = 16

and third proportion = 128

$\therefore \sqrt{x y}=16 \Rightarrow x y=256$

$\Rightarrow x=\frac{256}{y}$...(i)

and $\frac{y^{2}}{x}=128 \Rightarrow x=\frac{y^{2}}{128}$...(ii)

From (i) and (ii)

$\frac{256}{y}=\frac{y^{2}}{128}$

$\Rightarrow y^{3}=256 \times 128=32768$

$\Rightarrow y^{3}=(32)^{3} \Rightarrow y=32$

$\therefore x=\frac{256}{y}=\frac{256}{32}=8$

$\therefore$ Numbers are 8,32

Question 12

If q is the mean proportional between p and r, prove that:

$p^{2}-3 q^{2}+r^{2}=q^{4}\left(\frac{1}{p^{2}}-\frac{3}{q^{2}}+\frac{1}{r^{2}}\right)$

Sol :

q is mean proportional between p and r

q² = pr

L.H.S. $=p^{2}-3 q^{2}+r^{2}=p^{2}-3 p r+r^{2}$

$\mathrm{R.H.S.}=q^{4}\left(\frac{1}{p^{2}}-\frac{3}{q^{2}}+\frac{1}{r^{2}}\right)$

$=\left(q^{2}\right)^{2}\left(\frac{1}{p^{2}}-\frac{3}{q^{2}}+\frac{1}{r^{2}}\right)=(p r)^{2}\left(\frac{1}{p^{2}}-\frac{3}{q^{2}}+\frac{1}{r^{2}}\right)$

$=(p r)^{2}\left(\frac{1}{p^{2}}-\frac{3}{p r}+\frac{1}{r^{2}}\right)$

$=p^{2} r^{2}\left(\frac{r^{2}-3 p r+p^{2}}{p^{2} r^{2}}\right)=r^{2}-3 p r+p^{2}$

∴L.H.S=R.H.S

Question 13

If $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$ , prove that each ratio is 

(i) $\sqrt{\frac{3 a^{2}-5 c^{2}+7 e^{2}}{3 b^{2}-5 d^{2}+7 t^{2}}}$

(ii) $\left[\frac{2 a^{3}+5 c^{3}+7 e^{3}}{2 b^{3}+5 d^{3}+7 f^{3}}\right]^{\frac{1}{3}}$

Sol :

$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k(\mathrm{say})$

∴ a = k, c = dk, e = fk

(i) $\sqrt{\frac{3 a^{2}-5 c^{2}+7 e^{2}}{3 b^{2}-5 d^{2}+7 f^{2}}}$

$=\sqrt{\frac{3 b^{2} k^{2}-5 d^{2} k^{2}+7 f^{2} k^{2}}{3 b^{2}-5 d^{2}-7 f^{2}}}$

$=k \sqrt{\frac{3 b^{2}-5 d^{2}+7 f^{2}}{3 b^{2}-5 d^{2}+7 f^{2}}}=k$

hence proved

(ii) $\left[\frac{2 a^{3}+5 c^{3}+7 e^{3}}{2 b^{3}+5 d^{3}+7 f^{3}}\right]^{\frac{1}{3}}$

$=\left[\frac{2 b^{3} k^{3}+5 d^{3} k^{3}+7 f^{3} k^{3}}{2 b^{3}+5 d^{3}+7 f^{3}}\right]^{\frac{1}{3}}$

$=k\left[\frac{2 b^{3}+5 d^{3}+7 f^{3}}{2 b^{3}+5 d^{3}+7 f^{3}}\right]^{\frac{1}{3}}=k$

hence proved

Question 14

If $\frac{x}{a}=\frac{y}{b}=\frac{z}{c},$ prove that $\frac{3 x^{3}-5 y^{3}+4 z^{3}}{3 a^{3}-5 b^{3}+4 c^{3}}=\left(\frac{3 x-5 y+4 z}{3 a-5 b+4 c}\right)^{3}$

Sol :

$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k$( say )

x=a k, y=b k, z=c k

L.H.S. $=\frac{3 x^{3}-5 y^{3}+4 z^{3}}{3 a^{3}-5 b^{3}+4 c^{3}}$

$=\frac{3 a^{3} k^{3}-5 b^{3} k^{3}+4 c^{3} k^{3}}{3 a^{3}-5 b^{3}+4 c^{3}}$

$=\frac{k^{3}\left(3 a^{3}-5 b^{3}+4 c^{3}\right)}{3 a^{3}-5 b^{3}+4 c^{3}}=k^{3}$

R.H.S. $=\left(\frac{3 x-5 y+4 z}{3 a-5 b+4 c}\right)^{3}$

$=\left(\frac{3 a k-5 b k+4 c k}{3 a-5 b+4 c}\right)^{3}=\left(\frac{k(3 a-5 b+4 c)}{(3 a-5 b+4 c)}\right)^{3}$

$=(k)^{3}=k^{3}$

∴L.H.S=R.H.S

Question 15

If x : a = y : b, prove that

$\frac{x^{4}+a^{4}}{x^{3}+a^{3}}+\frac{y^{4}+b^{4}}{y^{3}+b^{3}}=\frac{(x+y)^{4}+(a+b)^{4}}{(x+y)^{3}+(a+b)^{3}}$

Sol :

$\frac{x}{a}=\frac{y}{b}=k(s a y)$

x = ak, y = bk

L.H.S. $=\frac{x^{4}+a^{4}}{x^{3}+a^{3}}+\frac{y^{4}+b^{4}}{y^{3}+b^{3}}$

$=\frac{a^{4} k^{4}+a^{4}}{a^{3} k^{3}+a^{3}}+\frac{b^{4} k^{4}+b^{4}}{b^{3} k^{3}+b^{3}}$

$=\frac{a^{4}\left(k^{4}+1\right)}{a^{3}\left(k^{3}+1\right)}+\frac{b^{4}\left(k^{4}+1\right)}{b^{3}\left(k^{3}+1\right)}$

$=\frac{a\left(k^{4}+1\right)}{k^{3}+1}+\frac{b\left(k^{4}+1\right)}{k^{3}+1}$

$=\frac{a\left(k^{4}+1\right)+b\left(k^{4}+1\right)}{k^{3}+1}=\frac{\left(k^{4}+1\right)(a+b)}{k^{3}+1}$

R.H.S. $=\frac{(x+y)^{4}+(a+b)^{4}}{(x+y)^{3}+(a+b)^{3}}$

$=\frac{(a k+b k)^{4}+(a+b)^{4}}{(a k+b k)^{3}+(a+b)^{3}}$

$=\frac{k^{4}(a+b)^{4}+(a+b)^{4}}{k^{3}(a+b)^{3}(a+b)^{3}}=\frac{(a+b)^{4}\left(k^{4}+1\right)}{(a+b)^{3}\left(k^{3}+1\right)}$

$=\frac{(a+b)\left(k^{4}+1\right)}{k^{3}+1}=\frac{\left(k^{4}+1\right)(a+b)}{k^{3}+1}$

∴L.H.S=R.H.S

Hence proved

Question 16

If $\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}$ prove that each ratio’s equal to :

$\frac{x+y+z}{a+b+c}$

Sol :

$\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}=k($ say $)$

x = k(b + c – a),

y = k(c + a – b),

z = k(a + b – c)

$\frac{x+y+z}{a+b+c}$

$=\frac{k(b+c-a)+k(c+a-b)+k(a+b-c)}{a+b+c}$

$=\frac{k(b+c-a+c+a-b+a+b-c)}{a+b+c}$

$=\frac{k(a+b+c)}{a+b+c}=k$

Hence proved

Question 17

If a : b = 9 : 10, find the value of

(i) $\frac{5 a+3 b}{5 a-3 b}$

(ii) $\frac{2 \alpha^{2}-3 b^{2}}{2 a^{2}+3 b^{2}}$

Sol :

a : b = 9 : 10

$\Rightarrow \frac{a}{b}=\frac{9}{10}$

(i) $\frac{5 a+3 b}{5 a-3 b}=\frac{\frac{5 a}{b}+\frac{3 b}{b}}{\frac{5 a}{b}-\frac{3 b}{b}}=\frac{\frac{5 a}{b}+3}{\frac{5 a}{b}-3}$

(dividing by b)

$=\frac{5 \times \frac{9}{10}+3}{5 \times \frac{9}{10}-3}$

$=\frac{\frac{9}{2}+3}{\frac{9}{2}-3}$

$=\frac{\frac{15}{2}}{\frac{3}{2}}$

(Substituting the value of $\frac{a}{b}$)

$=\frac{15}{2} \times \frac{2}{3}=5$

(ii) \frac{2 a^{2}-3 b^{2}}{2 a^{2}+3 b^{2}}$

$=\frac{\frac{2 a^{2}}{b^{2}}-\frac{3 b^{2}}{b^{2}}}{\frac{2 a^{2}}{b^{2}}-\frac{3 b^{2}}{h^{2}}}

(Dividing by $b^{2}$)

=\frac{2\left(\frac{a}{b}\right)^{2}-3}{2\left(\frac{a}{b}\right)^{2}+3}$

$=\frac{2\left(\frac{9}{10}\right)^{2}-3}{2\left(\frac{9}{10}\right)^{2}+3}

$=\frac{2 \times \frac{81}{100}-3}{2 \times \frac{81}{100}+3}$

$=\frac{\frac{81}{50}-3}{\frac{81}{50}+3}$

$=\frac{\frac{81-150}{50}}{\frac{81+150}{50}}$

$=\frac{-69}{50} \times \frac{50}{231}=\frac{-69}{231}=\frac{-23}{77}$

Question 18

If (3x² + 2y²) : (3x² – 2y²) = 11 : 9, find the value of  $\frac{3 x^{4}+25 y^{4}}{3 x^{4}-25 y^{4}}$

Sol :

$\frac{3 x^{4}+25 y^{4}}{3 x^{4}-25 y^{4}}=\frac{11}{9}$

Applying componendo and dividendo

$\frac{3 x^{2}+2 y^{2}+3 x^{2}-2 y^{2}}{3 x^{2}+2 y^{2}-3 x^{2}+2 y^{2}}=\frac{11+9}{11-9}$

$\Rightarrow \frac{6 x^{2}}{4 y^{2}}=\frac{20}{2}$

$\Rightarrow  \frac{3}{2} \frac{x^{2}}{y^{2}}=10$

$\Rightarrow \frac{x^{2}}{y^{2}}=10 \times \frac{2}{3}=\frac{20}{3}$

$\frac{3 x^{4}+25 y^{4}}{3 x^{4}-25 y^{4}}$

$=\frac{\frac{3 x^{4}}{y^{4}}+\frac{25 y^{4}}{y^{4}}}{\frac{3 x^{4}}{y^{4}}-\frac{25 y^{4}}{y^{4}}}$

$=\frac{3\left(\frac{x^{2}}{y^{2}}\right)^{2}+25}{3\left(\frac{x^{2}}{y^{2}}\right)^{2}-25}$

$=\frac{3 \times\left(\frac{20}{3}\right)^{2}+25}{3\left(\frac{20}{3}\right)^{2}-25}$

$=\frac{3 \times \frac{400}{9}+25}{3 \times \frac{400}{9}-25}$

$=\frac{\frac{400}{3}+\frac{25}{1}}{\frac{400}{3}-\frac{25}{1}}$

$=\frac{\frac{400+75}{3}}{\frac{400-75}{3}}$

$=\frac{475}{3} \times \frac{3}{325}=\frac{19}{13}$

Question 19

If $x=\frac{2 m a b}{a+b}$ , find the value of

$\frac{x+m a}{x-m a}+\frac{x+m b}{x-m b}$

Sol :

$x=\frac{2 m a b}{a+b}$

$\Rightarrow \frac{x}{m a}+\frac{2 b}{a+b}$

Applying componendo and dividendo

$\frac{x+m a}{x-m a}=\frac{2 b+a+b}{2 b-a-b}=\frac{3 b+a}{b-a}$...(i)

Again, $\frac{x}{m b}=\frac{2 a}{a+b}$

Applying componendo and dividendo,

$\frac{x+m b}{x-m b}=\frac{2 a+a+b}{2 a-a-b}=\frac{3 a+b}{a-b}$..(ii)

Adding (i) and (ii)

$\frac{x+m a}{x-m a} \cdot+\frac{x+m b}{x-m b}=\frac{3 b+a}{b-a}+\frac{3 a+b}{a-b}$

$=-\frac{3 b+a}{a-b}+\frac{3 a+b}{a-b}=\frac{-3 b-a+3 a+b}{a-b}$

$=\frac{2 a-2 b}{a-b}=\frac{2(a-b)}{a-b}=2$

Question 20

If $x=\frac{p a b}{a+b},$ prove that $\frac{x+p a}{x-p a}-\frac{x+p b}{x-p b}=\frac{2\left(a^{2}-b^{2}\right)}{a b}$

Sol :

$x=\frac{p a b}{a+b}$

$\Rightarrow \frac{x}{p a}+\frac{b}{a+b}$

Applying componendo and dividendo

$\frac{x+p a}{x-p a}=\frac{b+a+b}{b-a-b}=\frac{a+2 b}{-a}$..(i)

Again $\frac{x}{p b}=\frac{a}{a+b}$

Applying componendo and dividendo,

$\frac{x+p b}{x-p b}=\frac{a+a+b}{a-a-b}=\frac{2 a+b}{-b}$..(ii)

$\mathrm{L.H.S.}=\frac{x+p a}{x-p a}-\frac{x+p b}{x-p b}$

$=\frac{a+2 b}{-a}-\frac{2 a+b}{-b}$

$=\frac{a+2 b}{-a}+\frac{2 a+b}{b}$

$=\frac{a b+2 b^{2}-2 a^{2}-a b}{-a b}$

$=\frac{2 b^{2}-2 a^{2}}{-a b}$

$=\frac{-2 a^{2}+2 b^{2}}{-a b}$

$=\frac{-2\left(a^{2}-b^{2}\right)}{-a b}=\frac{2\left(a^{2}-b^{2}\right)}{a b}$

=R.H.S

Question 21

Find x from the equation $\frac{a+x+\sqrt{a^{2}-x^{2}}}{a+x-\sqrt{a^{2}-x^{2}}}=\frac{b}{x}$

Sol :

$\frac{a+x+\sqrt{a^{2}-x^{2}}}{a+x-\sqrt{a^{2}-x^{2}}}=\frac{b}{x}$

Applying componendo and dividendo,

$\frac{a+x+\sqrt{a^{2}-x^{2}}+a+x-\sqrt{a^{2}-x^{2}}}{a+x+\sqrt{a^{2}-x^{2}}-a-x+\sqrt{a^{2}-x^{2}}}$

$=\frac{b+x}{b-x}$

$\Rightarrow \frac{2(a+x)}{2 \sqrt{a^{2}-x^{2}}}=\frac{b+x}{b-x}$

$\Rightarrow \frac{a+x}{\sqrt{a^{2}-x^{2}}}=\frac{b+x}{b-x}$

Squaring both sides,

$\frac{(a+x)^{2}}{a^{2}-x^{2}}=\frac{(b+x)^{2}}{(b-x)^{2}}$

$\Rightarrow \frac{(a+x)^{2}}{(a+x)(a-x)}=\frac{(b+x)^{2}}{(b-x)^{2}}$

$\Rightarrow \frac{a+x}{a-x}=\frac{(b+x)^{2}}{(b-x)^{2}}$

Again applying componendo and dividendo,

$\frac{a+x+a-x}{a+x-a+x}=\frac{(b+x)^{2}+(b-x)^{2}}{(b+x)^{2}-(b-x)^{2}}$

$\Rightarrow \frac{2 a}{2 x}=\frac{2\left(b^{2}+x^{2}\right)}{4 b x}$

$\Rightarrow \frac{a}{x}=\frac{b^{2}+x^{2}}{2 b x}$

$\Rightarrow \frac{2 a}{2 x}=\frac{2\left(b^{2}+x^{2}\right)}{4 b x}$

$\Rightarrow \frac{a}{x}=\frac{b^{2}+x^{2}}{2 b x}$

$\Rightarrow x^{2}=2 a b-b^{2}$

$x=\sqrt{2 a b-b^{2}}$ 

Question 22

If $x=\frac{\sqrt[3]{a+1}+\sqrt[3]{a-1}}{\sqrt[3]{a+1}-\sqrt[3]{a-1}},$ prove that

$x^{3}-3 a x^{2}+3 x-a=0$

Sol :

$x=\frac{\sqrt[3]{a+1}+\sqrt[3]{a-1}}{\sqrt[3]{a+1}-\sqrt[3]{a-1}}$

Applying componendo and dividendo,

$\frac{x+1}{x-1}=\frac{\sqrt[3]{a+1}+\sqrt[3]{a-1}+\sqrt[3]{a+1}-\sqrt[3]{a-1}}{\sqrt[3]{a+1}+\sqrt[3]{a-1}-\sqrt[3]{a+1}+\sqrt[3]{a-1}}$

$\frac{x+1}{x-1}=\frac{2 \sqrt[3]{a+1}}{2 \sqrt[3]{a-1}}$

$\Rightarrow \frac{x+1}{x-1}=\frac{\sqrt[3]{a+1}}{\sqrt[3]{a-1}}$

Cubing both sides

$\frac{(x+1)^{3}}{(x-1)^{3}}=\frac{a+1}{a-1}$

Again applying componendo and dividendo,

$\frac{(x+1)^{3}+(x-1)^{3}}{(x+1)^{3}-(x-1)^{3}}=\frac{a+1+a-1}{a+1-a+1}$

$\Rightarrow \frac{2\left(x^{3}+3 x\right)}{2\left(3 x^{2}+1\right)}=\frac{2 a}{2}$

$\Rightarrow \frac{x^{3}+3 x}{3 x^{2}+1}=\frac{a}{1}$

$\Rightarrow x^{3}+3 x=3 a x^{2}+a$

$\Rightarrow x^{3}-3 a x^{2}+3 x-a=0$

Question 23

If $\frac{b y+c z}{b^{2}+c^{2}}=\frac{c z+a x}{c^{2}+a^{2}}=\frac{a x+b y}{a^{2}+b^{2}}$, prove that each of these ratio is equal to 

Sol :

$\frac{b y+c z}{b^{2}+c^{2}}=\frac{c z+\alpha x}{c^{2}+a^{2}}=\frac{a x+b y}{a^{2}+b^{2}}$

$=\frac{2(a x+b y+c z)}{2\left(a^{2}+b^{2}+c^{2}\right)}=\frac{a x+b y+c z}{a^{2}+b^{2}+c^{2}}$ (Adding)

Now $\frac{b y+c z}{b^{2}+c^{2}}=\frac{a x+b y+c z}{a^{2}+b^{2}+c^{2}}$

$\Rightarrow \frac{b y+c z}{a x+b y+c z}=\frac{b^{2}+c^{2}}{a^{2}+b^{2}+c^{2}}$ (By alternendo)

$\Rightarrow \frac{b y+c z-a x-b y-c z}{a x+b y+c z}$

$=\frac{b^{2}+c^{2}-a^{2}-b^{2}-c^{2}}{a^{2}+b^{2}+c^{2}}$

$\Rightarrow \frac{-a x}{a x+b y+c z}=\frac{-a^{2}}{a^{2}+b^{2}+c^{2}}$

$\Rightarrow \frac{x}{a x+b y+c z}=\frac{a}{a^{2}+b^{2}+c^{2}}$

$\Rightarrow \frac{x}{a}=\frac{a x+b y+c z}{a^{2}+b^{2}+c^{2}}$ ..(i)

Similarly we can prove that

$\frac{y}{b}=\frac{a x+b y+c z}{a^{2}+b^{2}+c^{2}}$..(ii)

and $\frac{z}{c}=\frac{a x+b y+c z}{a^{2}+b^{2}+c^{2}}$...(iii)

From (i), (ii) and (iii)

Hence $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$