Without actually calculating the cubes,find t | vedclass

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(D) We use the algebraic identity: If $x + y + z = 0$,then $x^3 + y^3 + z^3 = 3xyz$.Given expression: $(31)^3 + (-16)^3 + (-15)^3$.Let $x = 31$,$y = -

Vedclass · Accepted Answer

$22320$

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(D) We use the algebraic identity: If $x + y + z = 0$,then $x^3 + y^3 + z^3 = 3xyz$.Given expression: $(31)^3 + (-16)^3 + (-15)^3$.Let $x = 31$,$y = -16$,and $z = -15$.Check the sum: $x + y + z = 31 + (-16) + (-15) = 31 - 31 = 0$.Since the sum is $0$,the value of the expression is $3xyz$.Value $= 3 	imes (31) 	imes (-16) 	imes (-15)$.Value $= 3 	imes 31 	imes 240$.Value $= 93 	imes 240 = 22320$.

Without actually calculating the cubes,find the value of each of the following:
$(31)^{3} - (16)^{3} - (15)^{3}$

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Without actually calculating the cubes,find the value of each of the following:$(31)^{3} - (16)^{3} - (15)^{3}$