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Question Number 86684 by john santu last updated on 30/Mar/20

$$\mathrm{If}{(1+\mathrm{px}+{\mathrm{qx}}^2)}^8=1+8\mathrm{x}+{52\mathrm{x}}^2+{\mathrm{kx}}^3+...$$$$\mathrm{find}\mathrm{p},\mathrm{q}\mathrm{and}\mathrm{k}.$$

Answered by jagoll last updated on 30/Mar/20

Commented by john santu last updated on 30/Mar/20

$$\mathrm{wow}...\mathrm{super}$$

Answered by Rio Michael last updated on 30/Mar/20

$$\mathrm{my}\mathrm{try}.$$$$\mathrm{let}\mathrm{p}x+qx=X$$$$\Rightarrow {(1+px+{qx}^2)}^8={(1+X)}^8=1+8x+52x^2+{kx}^3$$$${(1+X)}^8=1+8X+\frac{8\left(7\right)}{2}{X}^2+...$$$$=1+8(px+{qx}^2)+28{(px+{qx}^2)}^2+...$$$$=1+8px+8{qx}^2+28p^2{x}^2+56{pqx}^3+...$$$$\mathrm{as}1=1$$$$8p=8\Rightarrow p=1$$$$8q+28p^2=52\Rightarrow q=3$$$$k=56pq+56\Rightarrow k=224$$

Commented by jagoll last updated on 30/Mar/20

$$\mathrm{good}..$$

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