Rewrite each expression as a sum or difference of terms. (i) (x+3)(x-3) (ii) (x^2-3)(x^2+3) (iii) (x^15+3)(x^15-3) (iv) (x-3)(x^2-9)(x+3) (v) (x^2+y^2)(x^2-y^2) (vi) (x^2+y^2)^2 (vii) (x-y)²(x+y)² (viii) (x-y)²(x²+y²)² (x+y)²

Answer:All the answers are given below.Step-by-step explanation:(i) (x+3)(x-3) =x² -3² =x² -9 {Since (a+b)(a-b) =a² -b²}(ii) (x²-3)(x²+3) =(x²)² -3² =[tex]x^{4} -9[/tex] {Since (a+b)(a-b) =a² -b²}(iii) [tex](x^{15} +3)(x^{15}-3) =(x^{15} )^{2}-3^{2}  =x^{30}  -9[/tex] (iv) (x-3)(x²-9)(x+3) =(x²-9)(x²-9)=(x²-9)²=[tex]x^{4}-18x^{2} +81[/tex]{Since, (a-b)² =a²-2ab+b²}(v) (x²+y²)(x²-y²) = (x²)² -(y²)² =[tex]x^{4}-y^{4}[/tex](vi) (x²+y²)² =[tex]x^{4}+2x^{2}  y^{2} +y^{4}[/tex] {Since (a+b)² =a²+2ab+b²}(vii) (x-y)²(x+y)² =(x²-y²)² =[tex]x^{4} -2x^{2} y^{2}+y^{4}[/tex] {Since (a-b)² =a²-2ab+b² and (a+b)(a-b) =a²-b²}(viii) (x-y)²(x²+y²)²(x+y)² =(x²+y²)²(x²-y²)² =[tex](x^{4} -y^{4} )^{2} =x^{8}-2x^{4}  y^{4} +y^{8}[/tex] (Answer)