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(B) We know the algebraic identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.If $x+y+z = 0$,then $x^{3}+y^{3}+z^{3} = 3xyz$.We n

Vedclass · Accepted Answer

$77760$

Vedclass · Answer

(B) We know the algebraic identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.If $x+y+z = 0$,then $x^{3}+y^{3}+z^{3} = 3xyz$.We need to find the value of $48^{3}-30^{3}-18^{3}$,which can be written as $48^{3}+(-30)^{3}+(-18)^{3}$.Let $x = 48$,$y = -30$,and $z = -18$.Check the sum: $x+y+z = 48 + (-30) + (-18) = 48 - 48 = 0$.Since the sum is $0$,we can use the identity $x^{3}+y^{3}+z^{3} = 3xyz$.Therefore,$48^{3}+(-30)^{3}+(-18)^{3} = 3 	imes 48 	imes (-30) 	imes (-18)$.Calculating the product: $3 	imes 48 = 144$,and $(-30) 	imes (-18) = 540$.$144 	imes 540 = 77760$.

Without actually calculating the cubes,find the value of $48^{3}-30^{3}-18^{3}$.

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