If x+y+z=9, then the value of (x-4)^3+(y-2)^3+(z-3)^3-3 (x'4)(y-2)(z-3) is
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now
let x-4 be a y-2 be b and z-3 be c
a+b+c=x-4+y-2+z-3=(x+y+z)-4-2-3=9-9=0
we know that
when a+b+c=0 then a^3+b^3+c^3-3abc=0
hence answer =0
let x-4 be a y-2 be b and z-3 be c
a+b+c=x-4+y-2+z-3=(x+y+z)-4-2-3=9-9=0
we know that
when a+b+c=0 then a^3+b^3+c^3-3abc=0
hence answer =0
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