Question

a. (x+y+4)²- (2x+3y-1)²

b. x¹⁶-1

Answer 1

a. \(\left(x+y+4\right)^2-\left(2x+3y-1\right)^2=\left(x+y+4-2x-3y+1\right)\left(x+y+4+2x+3y-1\right)=\left(5-x-2y\right)\left(3x+4y+3\right)\)b. \(x^{16}-1=\left(x^8\right)^2-1^2=\left(x^8-1\right)\left(x^8+1\right)=\left[\left(x^4\right)^2-1^2\right]\left(x^8+1\right)=\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\)