Question : If $3a^{2}= b^{2}\neq0$, then the value of $\frac{\left (a+b \right)^{3}-\left (a-b \right)^{3}}{\left (a+b \right)^{2}+\left (a-b \right)^{2}}$ is:
Option 1: $\frac{3b}{2}$
Option 2: $b$
Option 3: $\frac{b}{2}$
Option 4: $\frac{2b}{3}$
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Correct Answer: $\frac{3b}{2}$
Solution :
Expression = $\frac{(a + b)^3 - (a - b)^3 }{(a + b)^2 + (a - b)^2} $ and $3a^{2}= b^{2}\neq0$
Using identities: $(a + b)^3=a^3 + b^3 + 3a^2b + 3ab^2$ and $(a - b)^3=a^3 - b^3 - 3a^2b + 3ab^2$,
Now, $\frac{(a + b)^3 - (a - b)^3 }{(a + b)^2 + (a - b)^2}$
= $ \frac{(a^3 + b^3 + 3a^2b + 3ab^2) - (a^3 - b^3 - 3a^2b + 3ab^2)}{(a^{2} + b^{2} + 2ab) + (a^{2} + b^{2} - 2ab)}$
= $\frac{\left(a^{3} + b^{3} + 3a^{2}b + 3ab^{2} - a^{3} + b^{3} +3a^{2}b - 3ab^{2}\right)}{(a^2 + b^2 + a^2 + b^2)}$
= $\frac{6a^2b + 2b^3}{2(a^2 + b^2)}$
Since $3a^{2}= b^{2}\neq0$,
$\frac{(a + b)^3 - (a - b)^3 }{(a + b)^2 + (a - b)^2}$
= $\frac{2b(3a^{2} + b^2)}{2(a^2 + b^2)}$
= $\frac{b(b^2 + b^2)}{\frac{b^2}{3}+b^2}$
= $\frac{b \times 2b^{2}}{\frac{4b^2}{3}}$
= $\frac{3b}{2}$
Hence, the correct answer is $ \frac{3b}{2}$.
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