If satisfies x^2 + x + 1 = 0 and (1 +)^7 = A + B + C ^2, where A, B, C 0, then 5(3A - 2B - C) is equal to:

Given: satisfies ^2+ +1=0. Find: Express (1+)^7 as A+B +C ^2 and compute 5(3A-2B-C). From ^2+ +1=0, the roots are the non-real cube roots of unity. So we may take = where ^3=1 and 1+ + ^2=0. Using 1+ =- ^2, (1+)^7=(1+)^7=(- ^2)^7=- ^14 Since ^3=1, we have ^12=1, so ^14= ^2. Therefore, (1+)^7=- ^2 Now from 1+ + ^2=0, we get - ^2=1+. Hence, (1+)^7=1+ Comparing 1+ with A+B +C ^2 for =, A=1, B=1, C=0 Therefore, 5(3A-2B-C)=5 (3 1-2 1-0)=5(1)=5 So the correct option is A. The solution lists the final value as 5, even though the answer key is inconsistent.

If satisfies x^2 + x + 1 = 0 and (1 +)^7 = A + B + C ^2, where A,… — Solution | JEE Mathematics

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If $\alpha$ satisfies $x^2 + x + 1 = 0$ and $(1 + \alpha)^7 = A + B\alpha + C\alpha^2$ , where $A, B, C \ge 0$ , then $5(3A - 2B - C)$ is equal to:

A
$5$
B
$6$
C
$7$
D
$8$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: $\alpha$ satisfies $\alpha^2+\alpha+1=0$ .

Find: Express $(1+\alpha)^7$ as $A+B\alpha+C\alpha^2$ and compute $5(3A-2B-C)$ .

From $\alpha^2+\alpha+1=0$ , the roots are the non-real cube roots of unity. So we may take $\alpha=\omega$ where $\omega^3=1$ and $1+\omega+\omega^2=0$ .

Using $1+\omega=-\omega^2$ ,

(1+\alpha)^7=(1+\omega)^7=(-\omega^2)^7=-\omega^{14}

Since $\omega^3=1$ , we have $\omega^{12}=1$ , so $\omega^{14}=\omega^2$ . Therefore,

(1+\alpha)^7=-\omega^2

Now from $1+\omega+\omega^2=0$ , we get $-\omega^2=1+\omega$ . Hence,

(1+\alpha)^7=1+\omega

Identify Coefficients

Comparing $1+\omega$ with $A+B\alpha+C\alpha^2$ for $\alpha=\omega$ ,

A=1, \qquad B=1, \qquad C=0

Therefore,

5(3A-2B-C)=5\bigl(3\cdot1-2\cdot1-0\bigr)=5(1)=5

So the correct option is A.

The solution lists the final value as $5$ , even though the answer key is inconsistent.

Common mistakes

Treating $\alpha$ as an arbitrary complex number instead of a cube root of unity is incorrect. From $\alpha^2+\alpha+1=0$ , we must use the root-of-unity relation $1+\alpha+\alpha^2=0$ .
Reducing powers incorrectly is a common error. Since $\omega^3=1$ , powers must be reduced modulo $3$ , so $\omega^{14}=\omega^2$ , not $\omega$ or $1$ .
After obtaining $-\omega^2$ , stopping there is incomplete. It must be rewritten using $1+\omega+\omega^2=0$ as $1+\omega$ to match the form $A+B\alpha+C\alpha^2$ .

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Answer

Step-by-step solution

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