You need to use the binomial formula, such that:

`(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k) y^k`

You need to replace 3a for x, 4b for y and 5 for n, such that:

`(3a - 4b)^5 = 5C0 (3a)^5+5C1 (3a)^4*(-4b)^1+5C2 (3a)^3*(-4b)^2+5C3 (3a)^2*(-4b)^3 + 5C4 3a*(-4b)^4 + 5C5 (-4b)^5`

By definition, nC0 = nCn = 1, hence `5C0 = 5C5 = 1.`

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You need to use the binomial formula, such that:

`(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k) y^k`

You need to replace 3a for x, 4b for y and 5 for n, such that:

`(3a - 4b)^5 = 5C0 (3a)^5+5C1 (3a)^4*(-4b)^1+5C2 (3a)^3*(-4b)^2+5C3 (3a)^2*(-4b)^3 + 5C4 3a*(-4b)^4 + 5C5 (-4b)^5`

By definition, nC0 = nCn = 1, hence `5C0 = 5C5 = 1.`

By definition nC1 = nC(n-1) = n, hence `5C1 = 5C4 = 5.`

By definition `nC2 = (n(n-1))/2` , hence `5C2 = 5C3 = 10.`

`(3a- 4b)^5 = 243a^5 - 1620a^4*b+4320a^3*b^2-5760a^2*b^3 + 3840a*b^4 - 1024b^5`

**Hence, expanding the complex number using binomial theorem yields the simplified result `(3a- 4b)^5 = 243a^5 - 1620a^4*b+4320a^3*b^2-5760a^2*b^3 + 3840a*b^4 - 1024b^5.` **