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Question 14
Given that $$(5x - 3)^3 + (2x + 5)^3 + 27(4 - 3x)^3 = 9(3 - 5x)(2x + 5)(3x - 4)$$, then the value of $$(2x + 1)$$ is:
Solution
$$(5x - 3)^3 + (2x + 5)^3 + 27(4 - 3x)^3 = 9(3 - 5x)(2x + 5)(3x - 4)$$
=Â $$(5x - 3)^3 + (2x + 5)^3 + (12 - 9x)^3 = 3(3 - 5x)(2x + 5)(9x - 12)$$
= $$(5x - 3)^3 + (2x + 5)^3 + (12 - 9x)^3 = 3(5x - 3)(2x + 5)(12 - 9x)$$
If a + b +Â c = 0 then $$a^3 + b^3 + c^3 = 3abc$$
So, 5x - 3 + 2x + 5 + 12 - 9x = 0
-2x + 14 = 0
2x = 14
Now,
$$(2x + 1)$$ = 14 + 1 = 15
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