If a + b + c = 4 then find a^3 + b^3 + c^(3 ) = ?


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Question Number 60484 by Askash last updated on 21/May/19

$I f a + b + c = 4$ $t h e n f i n d$ $a^{3} + b^{3} + c^{3} = ?$
Commented by MJS last updated on 21/May/19
${ ((a+b+c=4)),((a^3 +b^3 +c^3 =x)) :} 2 equations in 4 unknown ⇒ parametric form with 2 parameters c=4−a−b x=a^3 +b^3 +(4−a−b)^3 with a, b ∈C that′s the perfect answer$
${\begin{cases} a + b + c = 4 \\ a^{3} + b^{3} + c^{3} = x \end{cases}$ $2 equations in 4 unknown \Rightarrow parametric$ $form with 2 parameters$ $c = 4 - a - b$ $x = a^{3} + b^{3} + {(4 - a - b)}^{3} with a, b \in C$ ${that}^{'} s the perfect answer$
Commented by MJS last updated on 21/May/19

$we get a unique solution only if a, b, c \in N^{★}$ $\Rightarrow a = 2 \land b = c = 1 \lor b = 2 \land a = c = 1 \lor c = 2 \land a = b = 1$ $\Rightarrow a^{3} + b^{3} + c^{3} = 10$
	Answered by tanmay last updated on 21/May/19

	$\frac{a + b + c}{3} ⩾ {(a b c)}^{\frac{1}{3}}$ $\frac{64}{27} ⩾ (a b c)$ $\frac{a^{3} + b^{3} + c^{3}}{3} ⩾ {(a^{3} b^{3} c^{3})}^{\frac{1}{3}}$ $a^{3} + b^{3} + c^{3} ⩾ 3 \times a b c$ $i f w e c o n s i d e r a b c = \frac{64}{27}$ $t h e n a^{3} + b^{3} + c^{3} ⩾ 3 \times \frac{64}{27}$ $a^{3} + b^{3} + c^{3} ⩾ \frac{64}{9}$
	Commented by Askash last updated on 21/May/19

	$I w a n t p e r f e c t a n s w e r .$
	Commented by mr W last updated on 21/May/19

	$h o w c a n y o u s a y A ⩾ C f r o m$ $A ⩾ B a n d B ⩽ C ?$
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