Find the value of frac{18^3 + 7^3 + 3 times 1 | vedclass

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Question

(A) The numerator is of the form $a^3 + b^3 + 3ab(a + b) = (a + b)^3$,where $a = 18$ and $b = 7$.Since $a + b = 18 + 7 = 25$,the numerator is $25^3$.T

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$1$

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(A) The numerator is of the form $a^3 + b^3 + 3ab(a + b) = (a + b)^3$,where $a = 18$ and $b = 7$.Since $a + b = 18 + 7 = 25$,the numerator is $25^3$.The denominator is of the form $(x + y)^6 = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} y^k$.Here,$x = 3$ and $y = 2$,so the denominator is $(3 + 2)^6 = 5^6$.Since $5^6 = (5^2)^3 = 25^3$,the denominator is $25^3$.Therefore,the value of the expression is $\frac{25^3}{25^3} = 1$.

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

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