Question

(1+i)^3+(1-i)^6 has the value of

Answer 1

your answer..........

Answer 2

Answer:

$(1+i)^3+(1-i)^6=10i-2$

Step-by-step explanation:

Given : Expression  $(1+i)^3+(1-i)^6$

To find : The value of the given expression?

Solution :

$(1+i)^3+(1-i)^6$

We apply the cube formula,

$(a+b)^3=a^3+b^3+3ab(a+b)$

$=1+i^3+3i(1+i)+(1-i^3-3i(1-i))^2$

$=1+i^3+3i+3i^2+(1-i^3-3i+3i^2))^2$

$=1-i+3i-3+(1+i-3i-3))^2$

$=2i-2+(-2i-2))^2$

Opening square formula,

$(a+b)^2=a^2+b^2+2ab$

$=2i-2+(-2i)^2+(-2)^2+2(-2i)(-2)$

$=2i-2-4+4+8i$

$=-2+10i$

Therefore, $(1+i)^3+(1-i)^6=10i-2$