Question : If $x= a-b, y=b-c, z=c-a$, then the numerical value of the algebraic expression $x^3+y^3+z^3-3xyz$ will be:
Option 1: $a + b + c$
Option 2: $0$
Option 3: $4(a + b + c)$
Option 4: $3abc$
Answer (1)
24th Jan, 2024
Correct Answer: $0$
Solution :
Given: $x= (a-b), y=(b-c), z=(c-a)$
$⇒x+y+z=a-b+b-c+c-a=0$
We know that, if $ x+y+z=0$, then $x^3+y^3+z^3 -3xyz =0$
Because,
$x^3+y^3+z^3 -3xyz = (x+y+z) (x^2+y^2+z^2-xy-yz-zx)$
Hence, the correct answer is $0$.
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