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Question Number 20488 by xing last updated on 27/Aug/17

Commented by ajfour last updated on 27/Aug/17

$$Questionis:$$$$Givena=-\sqrt{99}+\sqrt{999}+\sqrt{9999}$$$$b=\sqrt{99}-\sqrt{999}+\sqrt{9999}$$$$c=\sqrt{99}+\sqrt{999}-\sqrt{9999}$$$$Findthevalueof$$$$\frac{a^4}{(a-b)(a-c)}+\frac{b^4}{(b-c)(b-a)}+\frac{c^4}{(c-a)(c-b)}$$$$$$

Answered by ajfour last updated on 27/Aug/17

$$Leta=pc,b=qc;then$$$$\frac{a^4}{(a-b)(a-c)}+\frac{b^4}{(b-c)(b-a)}+\frac{c^4}{(c-a)(c-b)}$$$$=\frac{p^4{c}^4}{c^2(p-q)(p-1)}+\frac{q^4{c}^4}{c^2(q-1)(q-p)}$$$$+\frac{c^4}{c^2(1-p)(1-q)}$$$$=\frac{c^2}{(p-q)}[\frac{p^4}{p-1}-\frac{q^4}{q-1}+\frac{p-q}{(1-p)(1-q)}]$$$$=\frac{c^2}{(p-q)}\left[\frac{p^{4}q-p^4-{pq}^4+q^4+p-q}{(1-p)(1-q)}\right]$$$$=\frac{c^2}{(p-q)}\left[\frac{pq(p^3-q^3)-(p^4-q^4)+(p-q)}{(1-p)(1-q)}\right]$$$$=c^2\left[\frac{pq(p^2+q^2+pq)-(p^2+q^2)(p+q)+1}{(1-p)(1-q)}\right]$$$$=c^2\left[\frac{(p^2+q^2)(pq-p-q)+p^2{q}^2+1}{(1-p)(1-q)}\right]$$$$=c^2\left[\frac{(p^2+q^2)(p-1)(q-1)-p^2-q^2+p^2{q}^2+1}{(1-p)(1-q)}\right]$$$$=c^2\left[\frac{(p^2+q^2)(p-1)(q-1)+(p^2-1)(q^2-1)}{(1-p)(1-q)}\right]$$$$=c^2[p^2+q^2+(p+1)(q+1)]$$$$=c^2[p^2+q^2+pq+p+q+1]$$$$={\mathit{a}}^2+{\mathit{b}}^2+\mathit{a}\mathit{b}+\mathit{c}\mathit{a}+\mathit{b}\mathit{c}+{\mathit{c}}^2$$$$=\frac{1}{2}[{(\mathit{a}+\mathit{b})}^2+{(\mathit{b}+\mathit{c})}^2+{(\mathit{c}+\mathit{a})}^2]$$$$=\frac{1}{2}[4\times 9999+4\times 99+4\times 999]$$$$=2[9999+999+99]$$$$=18[1111+111+11]$$$$=18\times 1233)=9\times 2466$$$$=22194.\left(optionA\right)$$

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