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Multiplication of a Polynomial and Monomial

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Question

Expand ${(a + b)^{4} + (a - b)^{4}}$ and use it to ealuate
$(x^{2} + \sqrt{1 - x^{2}})^{4} + (x^{2} - \sqrt{1 - x^{2}})^{4}$

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Solution

We have
$(a + b)^{4} + (a - b)^{4} = [^{4} C_{0} a^{4} +^{4} C_{1} a^{3} b +^{4} C_{2} a^{2} b^{2} +^{6} C_{3} a b^{3} +^{4} C_{4} b^{4}] + [^{4} C_{0} a^{4} -^{4} C_{1} a^{3} b +^{4} C_{2} a^{2} b^{2} -^{4} c_{3} a b^{3} +^{4} C_{4} b^{4}] = 2 [^{4} C_{0} a^{4} +^{4} C_{2} a^{2} b^{2} +^{4} C_{4} b^{4}] Putting \sqrt{1 - x^{2}} = y, we get (x^{2} + \sqrt{1 - x^{2}})^{4} + (x^{2} - \sqrt{1 - x^{2}})^{4} (x^{2} + y)^{4} + (x^{2} - y)^{4} = 2 [^{4} C_{0} (x^{2})^{4} +^{4} C_{2} (x^{2})^{2} y^{2} +^{4} C_{4} y^{4}] = 2 [x^{6} + 6 x^{4} y^{2} + y^{4}] = 2 [x^{8} + 6 x^{4} (1 - x^{2}) + (1 - x^{2})^{2}] = 2 [x^{8} + (6 x^{4} - 6 x^{6}) + (1 - 2 x^{2} + x^{2})] = 2 [x^{8} - 6 x^{6} + 7 x^{4} - 2 x^{2} + 1] = 2 x^{8} - 12 x^{6} + 14 x^{4} - 4 x^{2} + 2$

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