Question

Find the value of $\frac{{18}^3+7^3+\mathrm{3.18.7.25}}{3^6+\mathrm{6.243.2}+\mathrm{15.81.4}+\mathrm{20.27.8}+\mathrm{15.9.16}+\mathrm{6.3.32}+64}$

• A. $\frac{1}{2}$
• B. $1$
• C. $2$
• D. $\frac{5}{2}$

Answer 1

Answer: (B)

$1$

Now if we look at the denominator, it is a simple binomial expansion of $(3+2{)}^6$
The numerator is ${18}^3+7^3+\mathrm{3.18.7.25}$
where,
${18}^3+7^3=(18+17{)}^3-\mathrm{3.18.17}(18+17)$ using $(a+b{)}^3=a^3+b^3+3ab(a+b)$
So the numerator becomes,
=$(25{)}^3-\mathrm{3.18.17}(25)+\mathrm{3.18.7.25}$
=$(5^2{)}^3$
=$5^6$
So the expression in the question will be,
$\frac{5^6}{5^6}=1$