If ((1 + i)^{6} + (1 - i)^{6})then (i^{2} = - 1)is equal to.... | MATH - INTEGRAL POWER OF IOTA, ALGEBRAIC OPERATIONS AND EQUALITY OF COMPLEX NUMBERS | SolveeIt

If $(1 + i)^{6} + (1 - i)^{6}$ then $i^{2} = - 1$ is equal to.

$i + i^{2} + i^{3} + ...$

$x = 3 + i$

$x^{3} - 3x^{2} - 8x + 15 =$

None of these

Answer

$x = 3 + i$

Explanation

$(x + y) + i( - x + y) = 1 - 3i$ ⇒ $x + y = 1$

$- x + y = - 3$

$\therefore$

Equating real and imaginary parts, we get

$x = 2,y = - 1$ and $(2, - 1)$

$\frac{(3 + 2i\sin\theta)(1 + 2i\sin\theta)}{(1 - 2i\sin\theta)(1 + 2i\sin\theta)}$ $(2x + 3y) + i( - 3x + 2y) = 4 + i$

Question: If \((1 + i)^{6} + (1 - i)^{6}\)then \(i^{2} = - 1\)is equal to....