Question 68

If $$(2x + 7)^3 + (2x + 8)^3 + (2x + 3)^3 = 3 (2x + 7) (2x + 8) (2x + 3)$$, then what is the value of $$x$$ ?

Solution

We knwo the identity

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

$$\Rightarrow a^3 + b^3 + c^3 = 3abc$$ only if $$(a+b+c)=0$$

So as in Given Question,

$$(2x + 7)^3 + (2x + 8)^3 + (2x + 3)^3 = 3 (2x + 7) (2x + 8) (2x + 3)$$ then

$$\therefore (2x + 7) + (2x + 8) + (2x + 3) = 0$$

$$\therefore 6x + 18 = 0$$

$$\therefore 6x = -18$$

$$\therefore x = \frac{-18}{6}$$

$$\therefore x = -3$$

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