Question

find the value of 1/(1cube + 2 cube + 3cube)power -3/2

Answer 1

here is ur answer dear
1/(1^3+2^3+3^3)^-3/2
=(1+8+27)^3/27
=(36)^3/2
=(36)^1/2+2/2
=(36)^1/2+1
=(36)^1/2×36
=root 36 ×36
=6×36
=216

Answer 2

Ans.

$\frac{1}{( { {1}^{3} + {2}^{3} + {3}^{3} })^{ - \frac{ 3}{2} } } \\ = \frac{1}{ ({1 + 8 + 27)}^{ - \frac{3}{2} } } \\ = \frac{1}{ {(36)}^{ - \frac{3}{2} } } \\ = {(36)}^{ \frac{3}{2} } \\ = {(36)}^{1 + \frac{1}{2} } \\ = {36}^{1} \times {36}^{ \frac{1}{2} } \\ = 36 \times \sqrt{36} \\ = 36 \times 6 \\ = 216$

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