Question

The value of $\frac{{23}^3+{27}^3}{{23}^2-23\times 27+{27}^2}$ is :

• A. 46
• B. 52
• C. 50
• D. 48

Answer 1

The correct option is C 50

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

We have, $\frac{{23}^3+{27}^3}{{23}^2-23\times 27+{27}^2}$

$=\frac{(23+27)({23}^2-23\times 27+{27}^2)}{({23}^2-23\times 27+{27}^2)}$ [Assume $23$ as ${}^{\prime }{a}^{\prime }$ and $27$ as ${}^{\prime }{b}^{\prime }$ ]

$=23+27$ [Cancelling $({23}^2-23\times 27+{27}^2)$ from numerator and denominator]

$=50.$