if a+b+c =0 and 1,w,w^2 are three cube roots of unity , then (a+bw+cw^2)^3+( a+bw^2+c^2)^3 is equal to
Answers
Answer:
w is a cube root of 1 , so
w^3 = 1
w^3 - 1 = 0
(w - 1)(w^2 + w + 1) = 0
w is not 1 , so w^2 + w + 1 = 0
Therefore w^2 + w = -1 .
Let X = a + bw + cw^2 and Y = a + bw^2 + cw .
So we want to prove X^3 + Y^3 = 27abc when a + b + c = 0 .
X + Y = a + bw + cw^2 + a + bw^2 + cw
= 2a + (b + c)(w^2 + w)
= 2a - (b + c)
= 3a or -3(b + c) <--- because a = -(b + c)
XY = (a + bw + cw^2)(a + bw^2 + cw)
= a^2 + ab*w^2 + acw + abw + b^2*w^3 + bc*w^2 + ac*w^2 + bc*w^4 + c^2*w^3
= a^2 + b^2 + c^2 + (ab + ac + bc)(w^2 + w) <--- because bc*w^4 = bcw
= a^2 + b^2 + c^2 - (ab + ac + bc)
= (a + b + c)^2 - 3(ab + ac + bc)
= -3(ab + ac + bc) <--- because a + b + c = 0
= -3((-b-c)b + (-b-c)c + bc)
= 3(b^2 + bc + c^2)
So X^3 + Y^3
= (X + Y)(X^2 - XY + Y^2)
= (X + Y)((X + Y)^2 - 3XY)
= (3a)((-3(b + c))^2 - 9(b^2 + bc + c^2))
= 3a(9(b^2 + 2bc + 9c^2) - 9(b^2 + bc + c^2))
= 3a(9bc)
= 27abc
So the equation is proved .
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