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Question: \(\alpha ,\beta \) are complex cube roots of unity and \(x=a+b\), \(y=a\alpha +b\beta \), \(z=a\beta...

α,β\alpha ,\beta are complex cube roots of unity and x=a+bx=a+b, y=aα+bβy=a\alpha +b\beta , z=aβ+bαz=a\beta +b\alpha then x3+y3+z3{{x}^{3}}+{{y}^{3}}+{{z}^{3}}=
A. 00
B. 3ab3ab
C. 3(a3−b3)3({{a}^{3}}-{{b}^{3}})
D. 3(a−b)33{{(a-b)}^{3}}

Explanation

Solution

Hint: Assume α=ω,β=ω2\alpha =\omega ,\beta ={{\omega }^{2}}. Substitute the values in x.yx.yand zz. Then add x+y+zx+y+z and simplify. Then use the formula x3+y3+z3=(x+y+z)(x2+y2+z2−xy−yz−zx)+3xyz{{x}^{3}}+{{y}^{3}}+{{z}^{3}}=(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)+3xyz. Then substitute the value of x+y+zx+y+z in the formula. You will get the answer. Try it.

Complete step-by-step answer:
x=a+bx=a+b
y=aω+bω2y=a\omega +b{{\omega }^{2}}
z=aω2+bωz=a{{\omega }^{2}}+b\omega

So x+y+z=a+b+aω+bω2+aω2+bωx+y+z=a+b+a\omega +b{{\omega }^{2}}+a{{\omega }^{2}}+b\omega
x+y+z=a(1+ω+ω2)+b(1+ω+ω2)x+y+z=a(1+\omega +{{\omega }^{2}})+b(1+\omega +{{\omega }^{2}})

We know 1+ω+ω2=01+\omega +{{\omega }^{2}}=0.
x+y+z=0x+y+z=0

We know the formula x3+y3+z3=(x+y+z)(x2+y2+z2−xy−yz−zx)+3xyz{{x}^{3}}+{{y}^{3}}+{{z}^{3}}=(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)+3xyz.
So herex+y+z=0x+y+z=0.

So x3+y3+z3=(0)(x2+y2+z2−xy−yz−zx)+3xyz{{x}^{3}}+{{y}^{3}}+{{z}^{3}}=(0)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)+3xyz
x3+y3+z3=3xyz{{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz
So we know the value ofx,yx,y and zz.

So substituting the values we get,
x3+y3+z3=3(a+b)(aω+bω2)(aω2+bω){{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3(a+b)(a\omega +b{{\omega }^{2}})(a{{\omega }^{2}}+b\omega )
Simplifying we get,

& {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3(a+b)({{a}^{2}}{{\omega }^{3}}+ab{{\omega }^{3}}+ab{{\omega }^{4}}+{{b}^{2}}{{\omega }^{3}}) \\\ & {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3({{a}^{3}}+{{b}^{3}}) \\\ \end{aligned}$$ $${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3({{a}^{3}}+{{b}^{3}})$$ So we get the value of $${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3({{a}^{3}}+{{b}^{3}})$$. Note: Read the question carefully. Also, while simplifying, don't make any mistake. Take utmost care of the sign. Do not jumble while simplifying. Solve it step by step. You must know the formula ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)+3xyz$.