Find the continued products:
(i) $(x + 1) (x - 1) (x^{2} + 1)$
(ii) $(x - 3) (x + 3) (x^{2} + 9)$
(iii) $(3 x - 2 y) (3 x + 2 y) (9 x^{2} + 4 y^{2})$
(iv) $(2 p + 3) (2 p - 3) (4 p^{2} + 9)$

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Solution

(i) $(x + 1) (x - 1) (x^{2} + 1) = {(x^{2}) - (1)^{2}} (x^{2} + 1) = (x^{2} - 1) (x^{2} + 1) = (x^{2})^{2} - (1)^{2} = x^{4} - 1$

(ii) $(x - 3) (x + 3) (x^{2} + 9) = {(x^{2}) - (3)^{2}} (x^{2} + 9) = (x^{2} - 9) (x^{2} + 9) = (x^{2})^{2} - (9)^{2} = x^{4} - 81$

(iii) $(3 x - 2 y) (3 x + 2 y) (9 x^{2} + 4 y^{2}) = {(3 x)^{2} - (2 y)^{2}} (9 x^{2} + 4 y^{2}) = (9 x^{2} - 4 y^{2}) (9 x^{2} + 4 y^{2}) = (9 x^{2})^{2} - (4 y^{2})^{2} = 81 x^{4} - 16 y^{4}$

(iv) $(2 p + 3) (2 p - 3) (4 p^{2} + 9) = {(2 p)^{2} - (3)^{2}} (4 p^{2} + 9) = (4 p^{2} - 9) (4 p^{2} + 9) - (4 p^{2})^{2} - (9)^{2} = 16 p^{4} - 81$

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Similar questions

(2 p + 3) (2 p - 3) (4 p^{2} + 9)

(3 x - 2 y) (3 x + 2 y) (9 x^{2} + 4 y^{2})

(x - 3) (x + 3) (x^{2} + 9)

\sqrt{3} x^{2} - 2 x + \frac{1}{2} = 0

x^{2} + \frac{1}{x^{2}} = 5

x - \frac{3}{x} = x^{2}

2 x^{2} - \sqrt{3 x} + 9 = 0

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x + \frac{1}{x} = 1

{(x + \frac{1}{x})}^{2} = 3 (x + \frac{1}{x}) + 4

x + \frac{1}{x} = x^{2}, x \neq 0

(x + 3) (x - 3) (x^{2} + 9)

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