Question

If $\frac{a^3+3a{b}^2}{3a^{2}b+b^3}=\frac{x^3+3x{y}^2}{3x^{2}y+y^3}$ then

• A. $bx=ay$
• B. $by=ax$
• C. $b^{2}y=a^{2}x$
• D. $b^{2}x=a^{2}y$

Answer 1

The correct option is A $bx=ay$

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

Use of compodendo and Devidendo

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

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

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

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

$\Rightarrow \left(a+b\right)\left(x+y\right)=\left(x+y\right)\left(a-b\right)$

$\Rightarrow ax-ay+bx-by=xa-xb+ay-yb$

$\Rightarrow bx+xb=ay+ay$

$\Rightarrow 2bx=2ay$

$\Rightarrow bx=ay$