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Question Number 60484 by Askash last updated on 21/May/19
If a + b + c = 4 then find a^3 + b^3 + c^(3 ) = ?
$${If}\:\:{a}\:+\:{b}\:+\:{c}\:=\:\mathrm{4} \\ $$$$ \\ $$$${then}\:{find} \\ $$$${a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} \:+\:{c}^{\mathrm{3}\:} =\:? \\ $$
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
$$\begin{cases}{{a}+{b}+{c}=\mathrm{4}}\\{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} ={x}}\end{cases} \\ $$$$\mathrm{2}\:\mathrm{equations}\:\mathrm{in}\:\mathrm{4}\:\mathrm{unknown}\:\Rightarrow\:\mathrm{parametric} \\ $$$$\mathrm{form}\:\mathrm{with}\:\mathrm{2}\:\mathrm{parameters} \\ $$$${c}=\mathrm{4}−{a}−{b} \\ $$$${x}={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +\left(\mathrm{4}−{a}−{b}\right)^{\mathrm{3}} \:\mathrm{with}\:{a},\:{b}\:\in\mathbb{C} \\ $$$$\mathrm{that}'\mathrm{s}\:\mathrm{the}\:\mathrm{perfect}\:\mathrm{answer} \\ $$
Commented by MJS last updated on 21/May/19
we get a unique solution only if a, b, c ∈N^★ ⇒ a=2∧b=c=1∨b=2∧a=c=1∨c=2∧a=b=1 ⇒ a^3 +b^3 +c^3 =10
$$\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:\mathrm{unique}\:\mathrm{solution}\:\mathrm{only}\:\mathrm{if}\:{a},\:{b},\:{c}\:\in\mathbb{N}^{\bigstar} \\ $$$$\Rightarrow\:{a}=\mathrm{2}\wedge{b}={c}=\mathrm{1}\vee{b}=\mathrm{2}\wedge{a}={c}=\mathrm{1}\vee{c}=\mathrm{2}\wedge{a}={b}=\mathrm{1} \\ $$$$\Rightarrow\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{10} \\ $$
Answered by tanmay last updated on 21/May/19
((a+b+c)/3)≥(abc)^(1/3) ((64)/(27))≥(abc) ((a^3 +b^3 +c^3 )/3)≥(a^3 b^3 c^3 )^(1/3) a^3 +b^3 +c^3 ≥3×abc if we consider abc=((64)/(27)) then a^3 +b^3 +c^3 ≥3×((64)/(27)) a^3 +b^3 +c^3 ≥((64)/9)
$$\frac{{a}+{b}+{c}}{\mathrm{3}}\geqslant\left({abc}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\frac{\mathrm{64}}{\mathrm{27}}\geqslant\left({abc}\right) \\ $$$$\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }{\mathrm{3}}\geqslant\left({a}^{\mathrm{3}} {b}^{\mathrm{3}} {c}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \geqslant\mathrm{3}×{abc} \\ $$$${if}\:{we}\:{consider}\:{abc}=\frac{\mathrm{64}}{\mathrm{27}} \\ $$$${then}\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \geqslant\mathrm{3}×\frac{\mathrm{64}}{\mathrm{27}} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \geqslant\frac{\mathrm{64}}{\mathrm{9}} \\ $$
Commented by Askash last updated on 21/May/19
I want perfect answer.
$${I}\:{want}\:{perfect}\:{answer}. \\ $$
Commented by mr W last updated on 21/May/19
how can you say A≥C from A≥B and B≤C ?
$${how}\:{can}\:{you}\:{say}\:{A}\geqslant{C}\:{from} \\ $$$${A}\geqslant{B}\:{and}\:{B}\leqslant{C}\:? \\ $$

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