\hspace{-16}$If $\mathbf{f(x)=x^{n+8}-10x^{n+6}+2x^{n+4}-10x^{n+2}+x^n+x^3-10x+1}$\\\\ Then Find $\mathbf{f(\sqrt{2}+\sqrt{3})=}\;,$ Where $\mathbf{n\in\mathbb{Z^{+}}}$
ans = 1+√2-√3
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1 Answers
1f(x)= xn[x8-10x6+2x4-10x2+1]+x3-10x+1
put x2 = t
→ f(x) = xn[t4-10t3+2t2-10t+1]+x3-10x+1
Notice that 'i' is a root of the eqn. inside the bracket. so it can be factorised as,
→ t4-10t3+2t2-10t+1 = (t2+1)(t2-10t+1)
→ 5+2 √6 = (√2+√3)2 is a root of the above eqn.
→ √2+√3 is a root of the original eqn. i.e. [x8-10x6+2x4-10x2+1
so the whole big term inside the bracket of the original equation cancels out.
hence f(√2+√3) = (√2+√3)3 -10(√2+√3) + 1
= 1+ √2-√3.
