If $$(x - 6)^3 + (x - 4)^3 + (x - 5)^3 = (3x - 15) (x - 4) (x - 6)$$, then what is the value of $$x$$?
Opening the LHS brackets we get :
3x^3-45x^2+231x-405 (1)(using (a-b)^3 =a^3-b^3-3a^2b+3ab^2)
Now opening RHS we get 3(x-5)(x-4)(x-6)Â
= 3(x^3-15x^2+74x-120)
= 3x^3-45x^2+222x-360Â Â (2)
Equating 1 and 2Â
we getÂ
9x=45
x=5
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