Question

10. If x, y and z are three different numbers, then
prove that:
x2 + y2 + z2 - xy - yz - zx is always positive.

11. Find :
(i) (a + b)(a + b)
(ii) (a + b)(a + b)(a + b)
(iii) (a - b) (a - b)(a - b) by using the result of part (ii)

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Answer 1

Answer:

10. Proved

11. See explanation

Step-by-step explanation:

10. $x^{2} + y^{2} + z^{2} - xy -yz - xz = \frac{1}{2} (x - y)^{2} + \frac{1}{2} (y - z)^{2} + \frac{1}{2} (z - x)^{2} > 0$ because square of number is always positive, and x,y,z are diffrent that means anyone expression is not equal 0.

11. $(i) (a + b)(a+b) = a^{2} + ab + ba + b^{2} = a^{2} + 2ab + b^{2} \\\\(ii) (a + b)(a+b)(a+b) = (a^{2} + 2ab + b^{2} )(a+b) = a^{3} + 2a^{2} b + ab^{2} + ba^{2} + 2ab^{2} + b^{3} = \\\\=a^{3} + 3a^{2} b + 3ab^{2} + b^{3} \\\\(iii) (a - b)(a - b)(a - b) = (a + (-b))(a + (-b))(a + (-b)) =\\\\ = a^{3} + 3a^{2}(-b) + 3a(-b)^{2} + (-b)^{3} = a^{3} - 3a^{2} b + 3ab^{2} - b^{3}$