Power of equations

Algebra Level 4

x 3 − 3 x y 2 = 54 y 3 − 3 x 2 y = 297 \large\begin{aligned}x^3-3xy^2=&\ 54\\y^3-3x^2y=&\ 297\end{aligned}

If x x and y y are real numbers that satisfy the system of equations above, find the value of x 2 + y 2 x^2+y^2 .


The answer is 45.

Ikkyu San by Ikkyu San , 23, Thailand

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2 solutions

Multiply the second equation by i i , then add both equations:

x 3 − 3 x 2 y i − 3 x y 2 + y 3 i = 54 + 297 i x^3-3x^2yi-3xy^2+y^3i=54+297i

Factor the left side:

( x − y i ) 3 = 54 + 297 i (x-yi)^3=54+297i

Take absolute value on both sides:

∣ ( x − y i ) 3 ∣ = ∣ 54 + 297 i ∣ |(x-yi)^3|=|54+297i|

Use the identity for any complex number z z : ∣ z n ∣ = ∣ z ∣ n |z^n|=|z|^n :

∣ x − y i ∣ 3 = ∣ 54 + 297 i ∣ |x-yi|^3=|54+297i|

Finally, use the definition of absolute value:

x 2 + y 2 3 = 5 4 2 + 29 7 2 ( x 2 + y 2 ) 3 = 5 4 2 + 29 7 2 x 2 + y 2 = 5 4 2 + 29 7 2 3 x 2 + y 2 = 2 2 × 2 7 2 + 1 1 2 × 2 7 2 3 x 2 + y 2 = 3 2 4 + 121 3 x 2 + y 2 = 9 125 3 x 2 + y 2 = 9 × 5 x 2 + y 2 = 45 \sqrt{x^2+y^2}^3=\sqrt{54^2+297^2} \\ (x^2+y^2)^3=54^2+297^2 \\ x^2+y^2=\sqrt[3]{54^2+297^2} \\ x^2+y^2=\sqrt[3]{2^2\times27^2+11^2\times27^2} \\ x^2+y^2=3^2\sqrt[3]{4+121}\\ x^2+y^2=9\sqrt[3]{125} \\ x^2+y^2=9\times 5 \\ x^2+y^2=\boxed{45}

2 Helpful 1 Interesting 0 Brilliant 0 Confused
Marc Ballon
Aug 23, 2015

Graphing the equations. Checking on what point did they intersect. (6,-3). 6^2 + (-3)^2=36+9=45

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Can u post an algebraic solution that doesn't involve graphing. Thanks great problem btw

Ashish Sacheti - 5 years, 9 months ago

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