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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

User Bemmu
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1 Answer

4 votes

Answer:

All the answers are given below.

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 =
x^(4) -9 {Since (a+b)(a-b) =a² -b²}

(iii)
(x^(15) +3)(x^(15)-3) =(x^(15) )^(2)-3^(2)  =x^(30)  -9

(iv) (x-3)(x²-9)(x+3) =(x²-9)(x²-9)=(x²-9)²=
x^(4)-18x^(2) +81{Since, (a-b)² =a²-2ab+b²}

(v) (x²+y²)(x²-y²) = (x²)² -(y²)² =
x^(4)-y^(4)

(vi) (x²+y²)² =
x^(4)+2x^(2)  y^(2) +y^(4) {Since (a+b)² =a²+2ab+b²}

(vii) (x-y)²(x+y)² =(x²-y²)² =
x^(4) -2x^(2) y^(2)+y^(4) {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 =
(x^(4) -y^(4) )^(2) =x^(8)-2x^(4)  y^(4) +y^(8) (Answer)

User Sean Christe
by
6.1k points
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