finding the value of $a$ and $b$

Question

If $a^3+12ab^2=679$ and $9a^2b+12b^3=978$, find $a^2-4ab+4b^2$.

Answer 1

$27(a^3+12ab^2)+18(9a^2b+12b^3)=27(679)+18(978)\implies (3a+6b)^3=33^3 \implies (a+2b)^3=11^3$

Similarly, $-27(a^3+12ab^2)+18(9a^2b+12b^3)=-27(679)+18(978)\implies (-a+2b)^3=(-3)^3$

Following is how $18,27$ are identified as the multipliers.

Clearly, $a\cdot b\ne 0$

$A^3(a^3+12ab^2)=(Aa)^3+3(Aa)(2Ab)^2$ and $18B^3(9a^2b+12b^3)=(6Bb)^3+3(6Bb)(3Ba)^2$

$A^3(a^3+12ab^2)+18B^3(9a^2b+12b^3)=A^3679+18B^3978$

If we put, $x=Aa,y^2=(2Ab)^2$ and $y=6Bb,x^2=(3Ba)^2,$ $A^3(a^3+12ab^2)+18B^3(9a^2b+12b^3)=(x+y)^3$

Comparing the values of $x^2,A^2=9B^2$ which is satisfied by comparing the values of $y^2$

So, $A=\pm 3B, A^3=\pm 27B^3,$

$\pm27B^3(a^3+12ab^2)+18B^3(9a^2b+12b^3)=(x+y)^3$

If $A=3B,$ and $x=Aa=3Ba, \{3B(a +2b)\}^3=(3B)^3679+18B^3978$

This is true for all the values of $B.$

If we take $B\ne 0, (a+2b)^3=1331\implies a+2b=11w_1$ where $w_1$ is a cube root of $1$

Similarly, for $A=-3B, (-a+2b)^3=-27\implies -a+2b=-3w_2$ where $w_2$ is a cube root of $1$

Now solve for $a,b$