Question

solve this plz........

Answer 1

Given : (1/3 + 3i)^3

It is in the form of (a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2

$= \ \textgreater \ ( \frac{1}{3} )^3 + (3i)^3 + 3( \frac{1}{3} )^2 (3i) + 3( \frac{1}{3})(3i)^2$

We know that i^3 = -i, i^2 = -1

$= \ \textgreater \ \frac{1}{27} - 27i + i - 9$

$= \ \textgreater \ ( \frac{1}{27} - 9) + i(-27 + 1)$

$= \ \textgreater \ \frac{-242}{27} - i26$



Hope this helps!