Question

a3+3ab2/3a2b+b3=x3+3xy2/3x2y+y3 show x/a=y/b​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\displaystyle \sf{ \frac{ {a}^{3} + 3 a{b}^{2} }{3 {a}^{2}b + {b}^{3} } = \: \frac{ {x}^{3} + 3 x{y}^{2} }{3 {x}^{2}y + {y}^{3} }}$

TO PROVE

$\displaystyle \sf{ \frac{x}{a} = \frac{y}{b} }$

PROOF

$\displaystyle \sf{ \frac{ {a}^{3} + 3 a{b}^{2} }{3 {a}^{2}b + {b}^{3} } = \: \frac{ {x}^{3} + 3 x{y}^{2} }{3 {x}^{2}y + {y}^{3} }}$

By the Componendo - Dividendo rule we get

$\displaystyle \sf{ \frac{ {a}^{3} + 3 a{b}^{2} + 3 {a}^{2}b + {b}^{3}}{ {a}^{3} + 3 a{b}^{2} - 3 {a}^{2}b - {b}^{3} } = \: \frac{ {x}^{3} + 3 x{y}^{2} + 3 {x}^{2}y + {y}^{3} }{{x}^{3} + 3 x{y}^{2} - 3 {x}^{2}y - {y}^{3} }}$

$\implies \: \displaystyle \sf{ \frac{ {(a + b)}^{3} }{ {(a - b)}^{3}} = \frac{ {(x + y)}^{3} }{ {(x - y)}^{3}} \: \: }$

$\implies \: \displaystyle \sf{ \frac{ {(a + b)}^{} }{ {(a - b)}^{}} = \frac{ {(x + y)}^{} }{ {(x - y)}^{}} \: \: }$

Again by Componendo Dividendo Rule

$\implies \: \displaystyle \sf{ \frac{ {(a + b + a + b)}^{} }{ {(a + b - a + b)}^{}} = \frac{ {(x + y + x - y)}^{} }{ {(x + y - x + y)}^{}} \: \: }$

$\implies \: \displaystyle \sf{ \frac{2a}{2b} = \frac{2x}{2y} }$

$\implies \: \displaystyle \sf{ \frac{a}{b} = \frac{x}{y} }$

$\implies \: \displaystyle \sf{ \frac{x}{a} = \frac{y}{b} }$

Hence proved

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Answer 2

Step-by-step explanation:

Hope this will help you..