Question 60

If $$(3x - 7)^3 + (3x - 8)^3 + (3x + 6)^3 = 3 (3x - 7) (3x - 8) (3x + 6)$$, then what is the value of $$x$$ ?

Solution

Let us consider,

(3x - 7) ,(3x - 8) and (3x + 6) = a, b, c respectively,

Hence ,

given $$a^3 +b^3 +c^3$$ = 3 a.b.c

Hence, if it is the case, Then a+b+c =0, therefore, (3x - 7 + 3x - 8 + 3x +6) = 0, hence 9x = 9, hence x = 1

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