The expression [x + (x3 - 1)½]5 + [x - (x3 - 1)½]5 is a polynomial of degree

1) 5

2) 6

3) 7

4) 8

Solution: (3) 7

\(\begin{array}{l}(a+b)^{5}+(a-b)^{5}\\ ={ }^{5} C_{0} a^{5}+{ }^{5} C_{1} a^{4} b+{ }^{5} C_{2} a^{3} b^{2}+{ }^{5} C_{3} a^{2} b^{3}+{ }^{5} C_{4} a b^{4}+{ }^{5} C_{5} b^{5}+{ }^{5} C_{0} a^{5}-{ }^{5} C_{1} a^{4} b +{ }^{5} C_{2} a^{3} b^{2}-{ }^{5} C_{3} a^{2} b^{3}+{ }^{5} C_{4} a b^{4}-{ }^{5} C_{5} b^{5}\\ =2\left[a^{5}+10 a^{3} b^{2}+5 a b^{4}\right]\\ \therefore \quad\left[x+\left(x^{3}-1\right)^{1 / 2}\right]^{5}+\left[x-\left(x^{3}-1\right)^{1 / 2}\right]^{5}\\ =2\left[x^{5}+10 x^{3}\left(x^{3}-1\right)+5 x\left(x^{3}-1\right)^{2}\right]\\ \text \ The \ given \ expression \ is \ a \ polynomial \ of \ degree \ 7.\end{array} \)

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