+50 express 16x^(4) -96x^(3)y +216x^(2)y^(2) -216xy^(3) +81y^(4) in the form of (ax+by)^n
+118608 \(16x^4 -96x^3y +216x^2y^2 -216xy^3 +81y^4=(ax+by)^n\\~\\ \text{If we just assume that this is actually possible then the answer is obvious}\\ n=4\\ a^4=16\;\; so\;\; a=\pm2\\ b^4=81\;\; so\;\; b=\pm3\\ \text{If a is positive b is negative and vise versa.}\\ (2x-3y)^4\qquad or \qquad (-2x+3y)^4 \)
But if you want to prove it is an identity then you have to go further.
\(16x^4 -96x^3y +216x^2y^2 -216xy^3 +81y^4=(ax+by)^n\\~\\ \text{Since the largest power of x and y is 4, }\;n=4\\ (ax+by)^4\\ =(ax)^4+4C1*(ax)^3(by)+4C2*(ax)^2(by)^2+4C3*(ax)^1(by)^3+(by)^4\\ =(ax)^4+4*(ax)^3(by)+6*(ax)^2(by)^2+4*(ax)^1(by)^3+(by)^4\\ a^4=16\quad so\;\;a=\pm2\\ b^4=81\quad so\;\;b=\pm3\\ =16x^4+4*(ax)^3(by)+6*4x^2*9y^2+4*(ax)^1(by)^3+81y^4\\ =16x^4+4a^3bx^3y+216x^2y^2+4ab^3xy^3+81y^4\\ \)
In order to get the negatives, when a is negative then b is positive and vise versa so\(4a^3b=-4*8*3=-96\\ 4ab^3=-4*2*27=-216\\~\\ (2x-3y)^4=16x^4-96x^3y+216x^2y^2-216xy^3+81y^4\\ and\\ (-2x+3y)^4=16x^4-96x^3y+216x^2y^2-216xy^3+81y^4\\\)
