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Let $$a=-\sqrt{99}+\sqrt{999}+\sqrt{9999}$$

$$b = \sqrt{99}-\sqrt{999}+\sqrt{9999}$$

$$c = \sqrt{99}+\sqrt{999}-\sqrt{9999}$$

Evaluate $$\frac{a^4}{(a-b)(a-c)}+\frac{b^4}{(b-c)(b-a)}+\frac{c^4}{(c-a)(c-b)}$$

Edited work :

$$\frac{a^4(b-c) + b^4(c-a)+c^4(a-b)}{(a-b)(a-c)(b-c)}$$

$$\frac{(a-b)(a-c)(b-c)(a^2+b^2+c^2+ab+ac+bc)}{(a-b)(a-c)(b-c)}$$

$$a^2+b^2+c^2+ab+ac+bc$$

$$a(a+b) +b(b+c) +c(c+a)$$

Ans : $$2(99+999+9999)=22194$$

user403160
  • 3,286

2 Answers2

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HINT: simplifying at first and you will get $${a}^{2}+ba+ac+{b}^{2}+bc+{c}^{2} $$ and now you can plug in the given values

Dr. Sonnhard Graubner
  • 95,283
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Since the given expression is symmetric in $a,b,c$, the result must be a symmetric second degree expression in $a,b,c$. Thus we can write $$\frac{a^4}{(a-b)(a-c)}+\frac{b^4}{(b-c)(b-a)}+\frac{c^4}{(c-a)(c-b)} = \lambda(a^2+b^2+c^2) + \mu(ab+bc+ca)$$ Putting $a=0,b=1,c=2$, we get $5\lambda+2\mu = 7$ and putting $a=0, b=1, c=-1$ we get $2\lambda - \mu = 1$. Thus $\lambda = \mu = 1$ and the given expression equals $a^2+b^2+c^2+ab+bc+ca$