Question 71

If $$a = 25, b = 15, c = -10$$, then the value of $$\frac{a^3 + b^3 + c^3 - 3 abc}{(a-b)^2 + (b-c)^2 + (c-a)^2}$$ is

Solution

Given a = 25, b = 15, c = -10

$$\frac{a^{3}+b^{3}+c^{3}-3abc}{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}$$ = $$\frac{(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)}{2(a^{2}+b^{2}+c^{2}-ab-bc-ca)}$$

= $$\frac{a+b+c}{2}$$

= $$\frac{25+15-10}{2}$$

= 15

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