Question

The value of $\frac{({18}^3+7^3+\mathrm{3.18.7.25})}{3^6+\mathrm{6.243.2}+\mathrm{15.81.4}+\mathrm{20.27.8}+\mathrm{15.9.16}+\mathrm{6.3.32}+64}$ is

• A. 1
• B. 5
• C. 25
• D. 100

Answer 1

The correct option is A 1

The numerator is of the form

$a^3$ + $b^3$ + 3ab(a + b) = $(a+b{)}^3$

$\therefore N=(18+7{)}^3={25}^3$

Denominator, $D=3^6+{}^6{C}_1{3}^5{2}^1+{}^6{C}_2{3}^4{2}^2+{}^6{C}_3{3}^3{2}^3+{}^6{C}_4{3}^2{2}^4+{}^6{C}_5{3}^1{2}^5+{}^6{C}_6{2}^6$

This is clearly the expansion of $(3+2{)}^6=5^6={25}^3$

$\therefore \frac{N}{D}=\frac{(25{)}^3}{(25{)}^3}=1$