Cubing the Cube Roots!?

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If x = a + b x=a+b
y = a w + b w 2 y=aw+bw^2
z = a w 2 + b w z= aw^2+bw

where w w is a complex Cube root of unity,
Then,

If f ( a , b ) = x 3 + y 3 + z 3 f(a,b) = x^3+y^3+z^3 ,

Evaluate f ( 3 , 4 ) f(3,4)


The answer is 273.

Anish Puthuraya by Anish Puthuraya , 24, India

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1 solution

Anish Puthuraya
Jan 19, 2014

We know,
1 + w + w 2 = 0 1+w+w^2 = 0

Therefore, it is clear by observation,
x + y + z = 0 x+y+z=0

Also,
x 3 + y 3 + z 3 − 3 x y z = ( x + y + z ) ( x 2 + y 2 + z 2 − x y − y z − z x ) x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)

Hence,
x 3 + y 3 + z 3 = 3 x y z x^3+y^3+z^3=3xyz
⇒ x 3 + y 3 + z 3 = 3 ( a 3 + b 3 ) \Rightarrow x^3+y^3+z^3=3\left(a^3+b^3\right)

Thus,
f ( a , b ) = 3 ( a 3 + b 3 ) f(a,b)=3\left(a^3+b^3\right)
⇒ f ( 3 , 4 ) = 273 \Rightarrow f(3,4) = \boxed{273}

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