Question

If 3x+y+z=0,show that 27x^3+y^3+z^3=9xyz

Answer 1

as 3x+y+z=0---(1)

also we know that,

"a³+b³+c³-3abc=(a+b+c)(a²+b²+c²+ab+bc+ca)"

so substituting the value of

'a' as 3x

'b' as y

'c' as z

so (3x)³+y³+z³-3(3x)yz=(3x+y+z)(3x²...+3xz)

so from (1)

27x³+y³+z³ - 9xyz=0

so,

27x³+y³+z³ = 9xyz

hence proved