Question
If $a + 2b + c = 0;$ then show that :$a^3+ 8b^3+ c^3= 6abc.$
✓
Answer
Given that $a + 2b + c = 0$ We know that
$\therefore a + 2b = - c ...(1)$
Cubing both sides of the above equation,
$a^3 + b^3 + c^3 = (a + b+ c) (a^2 + b^2 + c^2- ab - bc - ca) + 3abc$
If $a + b + c = 0,$ then,
$a^3 + b^3 + c^3= 3abc$
Here, $a= a, b = 2b$ and $c = c,$
Thus,
$a^3 + (2b)^3+ c^3 = 3(a) (2b) (c) = 6abc$
$\text{L.H.S} = \text{R.H.S}$
$\therefore a + 2b = - c ...(1)$
Cubing both sides of the above equation,
$a^3 + b^3 + c^3 = (a + b+ c) (a^2 + b^2 + c^2- ab - bc - ca) + 3abc$
If $a + b + c = 0,$ then,
$a^3 + b^3 + c^3= 3abc$
Here, $a= a, b = 2b$ and $c = c,$
Thus,
$a^3 + (2b)^3+ c^3 = 3(a) (2b) (c) = 6abc$
$\text{L.H.S} = \text{R.H.S}$
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