Question

Find \((x + 1)^6 - (x - 1)^6\). Hence or otherwise evaluate \((\sqrt2 + 1)^6 + (\sqrt2 - 1)^6.\)

Answer 1

Using Binomial Theorem, the expressions, \((x+ 1)^6\) and \((x -1)^6\), can be expanded as

\((x+1)^6\)= \(^6C_0 x^6+ ^6C_1 x^5\)\(+ ^6C_2 x^4 + ^6C_3 x^3\)\(+ ^6C_4 x^2 + ^6C_5 x + ^6C_6\)
\((x-1)^6\)= \(^6C_0 x^6 - ^6C_1 x^5 + ^6C_2 x^4\)\(- ^6C_3 x^3\)\(+ ^6C_4 x^2 - ^6C_5 x + ^6C_6\)
\((x+1)^6+(x-1)^6\)=\(2[ ^6C_0 x^6\) \(+ ^6C_2 x^4 +\) \(^6C_4 x^2+\) \(^6C_6]\)
=\(2[x^6+15x^4+15x^2+1]\)

\(\text{By putting } x = \sqrt2, \text{we obtain}\)

\((\sqrt2+1)^6+(\sqrt2-1)^6\) = \(2[ (\sqrt2)^6 +15(\sqrt2)^4 +15 (\sqrt2)^2+1]\)
=\(2(8+15×4+15 ×2+1)\)
=\(2(8+60+30+1)\)
=\(2(99)\)
=\(198\)

\(\text {Hence,} \,(\sqrt2 + 1)^6 + (\sqrt2 - 1)^6 = 198.\)