Solution of Cubes.

Algebra Level 3

How many triplets (x,y,z) in positive real numbers satisfy the following equations:

3 x 3 + 4 y 3 + 5 z 3 = 30 3x^{3}+4y^{3}+5z^{3} = 30 & 3 x 3 y 3 z 3 = 50 3x^{3}y^{3}z^{3} = 50


The answer is 1.

Yugesh Kothari by Yugesh Kothari , 22, India

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1 solution

Chew-Seong Cheong
Jul 27, 2015

It is given that:

{ 3 x 3 + 4 y 3 + 5 3 = 30 . . . ( 1 ) 3 x 3 y 3 z 3 = 50 . . . ( 2 ) \begin{cases} 3x^3+4y^3+5^3 = 30 & ... (1) \\ 3x^3 y^3 z^3 = 50 & ...(2) \end{cases}

Since x , y , z > 0 x,y,z>0 , we can use AM-GM inequality.

3 x 3 + 4 y 3 + 5 3 ≥ 3 60 x 3 y 3 z 3 3 ⇒ 30 ≥ 3 60 x 3 y 3 z 3 3 ⇒ 60 x 3 y 3 z 3 3 ≤ 10 60 x 3 y 3 z 3 ≤ 1000 3 x 3 y 3 z 3 ≤ 50 \begin{aligned} 3x^3+4y^3+5^3 & \ge 3 \sqrt[3]{60 x^3 y^3 z^3} \\ \Rightarrow 30 & \ge 3 \sqrt[3]{60 x^3 y^3 z^3} \\ \Rightarrow \sqrt[3]{60 x^3 y^3 z^3} & \le 10 \\ 60 x^3 y^3 z^3 & \le 1000 \\ 3 x^3 y^3 z^3 & \le 50 \end{aligned}

We note that from ( 1 ) (1) we have 3 x 3 y 3 z 3 ≤ 50 3 x^3 y^3 z^3 \le 50 and from ( 2 ) (2) we have 3 x 3 y 3 z 3 = 50 3 x^3 y^3 z^3 = 50 . This means that there is only 1 \boxed{1} triplet ( x , y , z ) ( x, y, z) that satifies the equations.

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Great application of AM-GM!

I'll just add in that the triplet is ( 10 3 3 , 5 2 3 , 2 3 ) (\sqrt[3]{\frac{10}{3}},\sqrt[3]{\frac{5}{2}},\sqrt[3]{2}) . Nice solution.

James Wilson - 3 years, 4 months ago

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