Question

equation .​Previous year IIT jee Question
Chapter :- complex number and quadratic​

Answer 1

EXPLANATION.

If ω is an imaginary cube roots.

⇒ (1 + ω - ω²)⁷.

As we know that,

⇒ 1 + ω + ω² = 0.

⇒ 1 + ω = - ω².

Put the values in the equation, we get.

⇒ (- ω² - ω²)⁷.

⇒ (- 2ω²)⁷.

⇒ (-2)⁷(ω)¹⁴.

⇒ (-2)⁷ (ω)¹²(ω)².

⇒ (-2)⁷(ω)².

⇒ - 128ω².

Option [D] is correct answer.

Answer 2

Solution:-

It is given that omega is cube root on unity and we need to find:-

$\large {\mathtt {(1 + \omega - {\omega}^{2}) }^{7}}$

Now we know that if omega is cube root of unity than

$\large {\mathtt {1 + \omega + {\omega}^{2} = 0}}$

From here

$\large {\mathtt {1 + \omega = - {\omega}^{2} }}$

Putting this in given Question we get

$\large {\mathtt {( - {\omega}^{2}- {\omega}^{2}) }^{7}}$

$\large {\mathtt {( - 2 {\omega}^{2}) }^{7}}$

$\large {\mathtt { - 128{\omega}^{14} }}$

As omega is cube root of unity than value of $\omega^3\: is \: 1$

$\large {\mathtt { - 128{\omega}^{14} =- 128 ({{\omega}^{3}})^{4} {\omega}^{2}}}$

$\large {\mathtt { - 128{\omega}^{14} =- 128 ({1})^{4} {\omega}^{2}}}$

$\large {\mathtt { - 128{\omega}^{14} =- 128 {\omega}^{2}}}$

$\huge \red{ \boxed{\mathtt{Answer=- 128 {\omega}^{2}}}}$