Polynomials Help (Relation of Roots) - Year 11 Maths Ext1 (1 Viewer)

[Scrambled]

I’ve managed to find the root to the cubic equation, but not entirely sure how to show pr^3 - q^3s = 0 (got a bit lost in the working out). Any help would be appreciated!

 

[liamkk112]

I’ve managed to find the root to the cubic equation, but not entirely sure how to show pr^3 - q^3s = 0 (got a bit lost in the working out). Any help would be appreciated!View attachment 43510View attachment 43508

well consider P(x) = px^3 +qx^2 + rx + s. Then, plug in a: you get p (-s/p) + q (-s/p)^(2/3) + r (-s/p)^(1/3) + s = 0. The s will cancel out, leaving you with
q ((s/p)^2)^(1/3) - r(s/p)^(1/3) = 0 by pulling out the minus sign. Thus q ((s/p)^2)^(1/3) = r (s/p)^(1/3) ;
cubing both sides gives q^3 (s/p)^2 = r^3 (s/p); so
q^3 (s/p) = r^3. from here the result is obvious

 

[Scrambled]

well consider P(x) = px^3 +qx^2 + rx + s. Then, plug in a: you get p (-s/p) + q (-s/p)^(2/3) + r (-s/p)^(1/3) + s = 0. The s will cancel out, leaving you with
q ((s/p)^2)^(1/3) - r(s/p)^(1/3) = 0 by pulling out the minus sign. Thus q ((s/p)^2)^(1/3) = r (s/p)^(1/3) ;
cubing both sides gives q^3 (s/p)^2 = r^3 (s/p); so
q^3 (s/p) = r^3. from here the result is obvious

ohhhh thank youu!