# $\alpha ,\beta $ are complex cube roots of unity and $x=a+b$, $y=a\alpha +b\beta $, $z=a\beta +b\alpha $ then ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}$=

A. $0$

B. $3ab$

C. $3({{a}^{3}}-{{b}^{3}})$

D. $3{{(a-b)}^{3}}$

Answer

Verified

363.9k+ views

Hint: Assume $\alpha =\omega ,\beta ={{\omega }^{2}}$. Substitute the values in $x.y$and $z$. Then add $x+y+z$ and simplify. Then use the formula ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)+3xyz$. Then substitute the value of $x+y+z$ in the formula. You will get the answer. Try it.

Complete step-by-step answer:

$x=a+b$

$y=a\omega +b{{\omega }^{2}}$

$z=a{{\omega }^{2}}+b\omega $

So \[x+y+z=a+b+a\omega +b{{\omega }^{2}}+a{{\omega }^{2}}+b\omega \]

\[x+y+z=a(1+\omega +{{\omega }^{2}})+b(1+\omega +{{\omega }^{2}})\]

We know \[1+\omega +{{\omega }^{2}}=0\].

\[x+y+z=0\]

We know the formula ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)+3xyz$.

So here\[x+y+z=0\].

So ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=(0)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)+3xyz$

${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz$

So we know the value of$x,y$ and $z$.

So substituting the values we get,

${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3(a+b)(a\omega +b{{\omega }^{2}})(a{{\omega }^{2}}+b\omega )$

Simplifying we get,

\[\begin{align}

& {{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{align}\]

\[{{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3({{a}^{3}}+{{b}^{3}})\]

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 ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)+3xyz$.

Complete step-by-step answer:

$x=a+b$

$y=a\omega +b{{\omega }^{2}}$

$z=a{{\omega }^{2}}+b\omega $

So \[x+y+z=a+b+a\omega +b{{\omega }^{2}}+a{{\omega }^{2}}+b\omega \]

\[x+y+z=a(1+\omega +{{\omega }^{2}})+b(1+\omega +{{\omega }^{2}})\]

We know \[1+\omega +{{\omega }^{2}}=0\].

\[x+y+z=0\]

We know the formula ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)+3xyz$.

So here\[x+y+z=0\].

So ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=(0)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)+3xyz$

${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz$

So we know the value of$x,y$ and $z$.

So substituting the values we get,

${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3(a+b)(a\omega +b{{\omega }^{2}})(a{{\omega }^{2}}+b\omega )$

Simplifying we get,

\[\begin{align}

& {{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{align}\]

\[{{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3({{a}^{3}}+{{b}^{3}})\]

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 ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)+3xyz$.

Last updated date: 03rd Oct 2023

â€¢

Total views: 363.9k

â€¢

Views today: 6.63k

Recently Updated Pages

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

One cusec is equal to how many liters class 8 maths CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE