Question 72

If a = 299, b = 298, c = 297 then the value of $$2a^{3} + 2b^{3} + 2c^{3}- 6abc$$ is

Solution

$$2a^{3} + 2b^{3} + 2c^{3}- 6abc$$,

Taking 2 common , and using , $$(a^3 + b^3 + c^3 - 3abc) = (a + b +c)(a^2 + b^2 +c^2 - ab - bc -ca)$$

= $$2(a + b +c)(a^2 + b^2 + c^2 - ab - bc - ca)$$

=$$2(894)(89401 + 88804 + 88209 - 89102 - 88506 - 88803)$$

= $$2 \times 894 \times 3$$

=5364

So, the answer would be option c)5364.

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