Question

The value of $\frac{{18}^3+7^3+3\cdot 18\cdot 7\cdot 25}{3^6+6\cdot 243\cdot 2+15\cdot 181\cdot 4+20\cdot 27\cdot 8+15\cdot 9\cdot 16+6\cdot 3\cdot 32+64}$

• A. $10$
• B. $1$
• C. $2$
• D. $20$

Answer 1

The correct option is A $10$

$\frac{{18}^3+7^3+3\cdot 18\cdot 7\cdot 25}{3^6+6\cdot 243\cdot 2+15\cdot 81\cdot 4+20\cdot 27\cdot 8+15\cdot 9\cdot 16+16\cdot 3\cdot 32+64}$

$=\frac{{18}^3+7^3+3\cdot 18\cdot 7(18+7)}{{}^6{C}_0.3^6{+}^6{C}_1.3^5.2{+}^6{C}_2.3^4.2^2{+}^6{C}_3.3^3.2^3{+}^6{C}_4.3^2.2^4{+}^6{C}_5.3.2^5{+}^6{C}_6.2^6}$

$=\frac{{(18+7)}^3}{{(3+2)}^6}[\because {(a+b)}^3=a^3+b^3+3ab(a+b);{(a+b)}^n{=}^n{C}_0.a^n{+}^n{C}_1.a^{n-1}{b}^1{+}^n{C}_{n-1}.a{b}^{n-1}{+}^n{C}_n.b^n]$

$=\frac{{25}^3}{5^6}=1$