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x^3+y^3+z^3 = (x+y+z)(x^2+y^2+z^2) - (xy+xz+yz)(x+y+z) + (xyz)(3) = (x+y+z)(x^2+y^2+z^2-[xy+xz+yz]) + 3xyz by factorising the first 2 terms. We know that (x-y)^2>=0 for all real x and y. It follows that x^2-2xy+y^2>=0, that is x^2+y^2>=2xy. Similarly, (x-z)^2>=0, that is x^2+z^2>=2xz and...

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Solution to question 2 attached

Solution attached - hope it makes sense

1. By the factor theorem, if (x+1) and (x-2) are factors, P(-1)=0 and P(2)=0 P(-1) = -1 + a - b + 2 = 0 Therefore, a - b = -1 ...... (1) P(2) = 8 + 4a + 2b + 2 = 0 Therefore, 2a + b = -5 ........ (2) (1) + (2): 3a = -6 Therefore, a = -2 Sub this in (1): -2 - b = -1, Therefore, b = -1...