Question

If the value of $\left(\frac{a^3-b^3}{a^3+b^3}\right)$, when $a=7$ and $b=2$ is $\frac{335}{p}$, then $p$ is $321$.

• A. 351
• B. False

Answer 1

The correct option is B False

$a^3-b^3=(a-b)(a^2+ab+b^2)$

$a^3+b^3=(a+b)(a^2-ab+b^2)$

Putting the values of $a$ and $b$, we get the value of $p$ as $351$.

Alternatively, we can observe that since $a$ and $b$ are positive, so the denominator has to be greater than the numerator. Thus, $a^3+b^3$ will be greater than $a^3-b^3$.