Question

if a = (x+y)^3 - (x-y)^3 and b = 3x^2+y^2 then what is A/B

Answer 1

Here,

$\begin{tabular}{ |c| } </p><p> \hline a= (x+y)^3 - (x-y)^3</p><p> \\ b = 3x^2+y^2 \\ </p><p> \hline</p><p>\end{tabular}$

To find : a/b

So, Simplifying the value of a,

$\to \tt{}a = (x+y)^3 - (x-y)^3$

• a³ - b³ = (a-b)(a²+b²+ab)

$\small \boxed{ \rm{applying \: \: above \: \: identity}} \\ \small \to \bf{} \{ (x + y) - (x - y)\} \{ {(x + y)}^{2} + {(x - y)}^{2} + (x + y)(x - y)\} \\ \small \to \bf{} \{ \cancel{x }+ y - \cancel{x} + y\} \{ {x}^{2} + \cancel{{y}^{2}} + \cancel{2xy }+ {x}^{2} + {y}^{2} -\cancel{ 2xy} + {x}^{2} - \cancel{{y}^{2} }\}\\ \small \boxed{ \rm{}cancelling \: \: like \: \: terms}\\ \small \to \bf{} \orange{2y \{ {3x}^{2} + {y}^{2} \} }$

Now,

$\huge \tt \implies \: \frac{a}{b} \: \\ \\ \large \rm \to \: \frac{2y( \cancel{{3x}^{2} + {y}^{2}})}{ \cancel{{3x}^{2} + {y}^{2} }} \\ \\ \huge \rm \to \: \red{ 2y}$

Thus,

• a/b = 2y