If $a+b+c=0$ ,show that, $(a+bomega+comega^2)^3+(a+bomega^2+comega)^3=27abc$ ?

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Answer 1 · 2018-05-29T10:03:07

$a + b + c = 0$

$LHS = (a + bω + cω²)³ + (a + bω² + cω)³$

If we consider $A = (a + bω + cω²)$ and $B = (a + bω² + cω)$

Adding the two.

$color(magenta)(A + B) = 2a + b(omega+omega^2) + c(omega^2 +omega)$
$=> 2a-b-c = 3a-a-b-c = 3a-0 = color(magenta)(3a)$ $"as " [ω² + ω + 1 = 0]$

Multiplying the two

$color(magenta)(AB) = (a + b + c)² - 3(ab + bc + ca) = color(magenta)(- 3(ab + bc + ca)$

As we know, $(A+B)^3 = A^3+B^3 +3AB(A+B)$

$P => A^3+B^3 = (A + B)³ - 3AB(A + B)$
$= 27a³ + 27a(ab + bc + ca)$
$= 27a(a² + ab + ac + bc) = 27a(a + b)(a + c) = 27a(-c)(-b) = 27abc$

Ifa+b+c=0a+b+c=0,show that,(a+bomega+comega^2)^3+(a+bomega^2+comega)^3=27abc(a+bomega+comega^2)^3+(a+bomega^2+comega)^3=27abc?