Question

(1+I)^3 + (1 - I)^6 has value of​

Answer 1

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

Answer 2

Answer:

10i - 2

Step-by-step explanation:

Given Equation is (1 + i)³ + (1 - i)⁶

(i)

(1 + i)³

= 1³ + 3 * 1² * i + 3 * 1 * i² + i³

= 1 + 3i + 3i² + i³

= 1 + 3i + 3(-1) + (-i)

= 1 + 3i - 3 - i

= 2i - 2

(ii)

(1 - i)⁶

= {(1 - i)²}³

= (1 + i² - 2i)³

= (1 - 1 - 2i)³

= -8i³

= -8(-i)

= 8i

Now,

(1 + i)³ + (1 - i)⁶

= 2i - 2 + 8i

= 10i - 2

Hope it helps!