Find $ (a + b)^4 - (a - b)^4.$ Hence, evaluate $( \sqrt3 + \sqrt2)^4 - (\sqrt 3 - \sqrt 2)^4$.

Question

$\begin{array}{1 1} 40 \sqrt 6 \\ 20 \sqrt 6 \\ 40 \sqrt 3 \\ 20 \sqrt 3 \end{array} $

Answer 1

Toolbox:

• Express the given number as the sum or difference of two numbers whose powers are easier to evaluate. Then use binomial theorem as follows:
• $(a+b)^n = \large \sum \limits_{k=0}^{n}\; $$^n \large C$$_k \; a^{n-k}b^k =\; ^n \large C$$_0\;a^nb^0 +\; ^n \large C$$_1\;a^{n-1}b^1+.... ^n\large C$$_n\;a^0b^n$, where $b^0 = 1 = a^{n-n}$

We can now use Binomial Theorem to evaluate $ (a + b)^4 – (a – b)^4$

$(a+b)^4 = ^4\large C$$_0\;a^nb^0 + ^4\large C$$_1\;a^{4-1}b^1+....$

$(a-b)^4 = ^4\large C$$_0\;a^nb^0 - ^4\large C$$_1\;a^{4-1}b^1+....$

$ (a + b)^4 – (a – b)^4 = ^4\large C$$_0\;a^nb^0 + ^4\large C$$_1\;a^{4-1}b^1+.... ^4\large C$$_4\;a^0b^4 - (^4\large C$$_0\;a^nb^0 - ^4\large C$$_1\;a^{4-1}b^1 + ^4\large C$$_4\;a^{4-4}b^4$)

$\qquad = 2 (\;^4\large C$$_1\;a^3b+\;^4\large C$$_3\;ab^3)$

$\qquad = 2 (4a^3b+4ab^3)$

$\qquad = 8ab(a^2+b^2)$

Therefore, $( \sqrt3 + \sqrt2)^4– (\sqrt 3 –\sqrt 2)^4 = 8 \sqrt 3 \sqrt 2 (\sqrt 3^2 + \sqrt 2^2)$

$\qquad = 8 \times \sqrt 6 \times 5 = 40 \sqrt 6 \approx 97.98$