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Question Number 48208 by naka3546 last updated on 20/Nov/18

f(x) + (x + 1)^3 = 2f(x + 1) f(10) = ?

$${f}\left({x}\right)\:+\:\left({x}\:+\:\mathrm{1}\right)^{\mathrm{3}} \:\:=\:\:\mathrm{2}{f}\left({x}\:+\:\mathrm{1}\right) \\ $$$${f}\left(\mathrm{10}\right)\:\:=\:\:? \\ $$

Answered by MJS last updated on 20/Nov/18

f(x)=ax^3 +bx^2 +cx+d f(x)+(x+1)^3 = =(a+1)x^3 +(b+3)x^2 +(c+3)x+(d+1) 2f(x+1)= =2ax^3 +2(3a+b)x^2 +2(3a+2b+c)x+2(a+b+c+d) comparing the constants a+1=2a b+3=2(3a+b) c+3=2(3a+2b+c) d+1=2(a+b+c+d) ⇒ a=1, b=−3, c=9, d=−13 f(x)=x^3 −3x^2 +9x−13 f(10)=777

$${f}\left({x}\right)={ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d} \\ $$$${f}\left({x}\right)+\left({x}+\mathrm{1}\right)^{\mathrm{3}} = \\ $$$$=\left({a}+\mathrm{1}\right){x}^{\mathrm{3}} +\left({b}+\mathrm{3}\right){x}^{\mathrm{2}} +\left({c}+\mathrm{3}\right){x}+\left({d}+\mathrm{1}\right) \\ $$$$\mathrm{2}{f}\left({x}+\mathrm{1}\right)= \\ $$$$=\mathrm{2}{ax}^{\mathrm{3}} +\mathrm{2}\left(\mathrm{3}{a}+{b}\right){x}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{3}{a}+\mathrm{2}{b}+{c}\right){x}+\mathrm{2}\left({a}+{b}+{c}+{d}\right) \\ $$$$\mathrm{comparing}\:\mathrm{the}\:\mathrm{constants} \\ $$$${a}+\mathrm{1}=\mathrm{2}{a} \\ $$$${b}+\mathrm{3}=\mathrm{2}\left(\mathrm{3}{a}+{b}\right) \\ $$$${c}+\mathrm{3}=\mathrm{2}\left(\mathrm{3}{a}+\mathrm{2}{b}+{c}\right) \\ $$$${d}+\mathrm{1}=\mathrm{2}\left({a}+{b}+{c}+{d}\right) \\ $$$$\Rightarrow\:{a}=\mathrm{1},\:{b}=−\mathrm{3},\:{c}=\mathrm{9},\:{d}=−\mathrm{13} \\ $$$${f}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{9}{x}−\mathrm{13} \\ $$$${f}\left(\mathrm{10}\right)=\mathrm{777} \\ $$

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