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(B) Let $a = 21$,$b = 15$,and $c = -36$.First,calculate the sum of $a, b,$ and $c$:$a + b + c = 21 + 15 + (-36) = 36 - 36 = 0$.We know the algebraic i

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$-34020$

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(B) Let $a = 21$,$b = 15$,and $c = -36$.First,calculate the sum of $a, b,$ and $c$:$a + b + c = 21 + 15 + (-36) = 36 - 36 = 0$.We know the algebraic identity: If $a + b + c = 0$,then $a^{3} + b^{3} + c^{3} = 3abc$.Substituting the values into the identity:$(21)^{3} + (15)^{3} + (-36)^{3} = 3 	imes (21) 	imes (15) 	imes (-36)$.Calculating the product:$3 	imes 21 = 63$$15 	imes (-36) = -540$$63 	imes (-540) = -34020$.Therefore,the value is $-34020$.

Without actually calculating the cubes,find the value of each of the following:
$(21)^{3} + (15)^{3} + (-36)^{3}$

Similar Questions

For the polynomial $\frac{x^{3}+2 x+1}{5}-\frac{7}{2} x^{2}-x^{6},$ write:
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Without actually calculating the cubes,find the value of each of the following:$(21)^{3} + (15)^{3} + (-36)^{3}$