Question

The value of $\frac{{43}^3-{27}^3}{{43}^2+43\times 27+{27}^2}$ is :

• A. 43
• B. 27
• C. 16
• D. 32

Answer 1

Answer: (C)

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

Now,

$\frac{{43}^3-{27}^3}{{43}^2+43\times 27+{27}^2}$

$=\frac{(43-27)({43}^2+43\times 27+{27}^2)}{({43}^2+43\times 27+{27}^2)}$ [Since, $a^3-b^3=(a^2+ab+b^2)(a-b)$]

$=43-27$

$=16$

$\therefore \frac{{43}^3-{27}^3}{{43}^2+43\times 27+{27}^2}=16$

Hence, Option D is correct.