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Question Number 156419 by cortano last updated on 11/Oct/21

x is positive integer number   can you check if Q=((((x+2)^4 −x^4 ))^(1/3)   is a natural number

$$\mathrm{x}\:\mathrm{is}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{number}\: \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{check}\:\mathrm{if}\:\mathrm{Q}=\sqrt[{\mathrm{3}}]{\left(\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{4}} −\mathrm{x}^{\mathrm{4}} \right.} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{natural}\:\mathrm{number} \\ $$

Commented by mr W last updated on 11/Oct/21

x=2:  Q=((4^4 −2^4 ))^(1/3) =((240))^(1/3) =2((30))^(1/3) ∉N

$${x}=\mathrm{2}: \\ $$$${Q}=\sqrt[{\mathrm{3}}]{\mathrm{4}^{\mathrm{4}} −\mathrm{2}^{\mathrm{4}} }=\sqrt[{\mathrm{3}}]{\mathrm{240}}=\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{30}}\notin{N} \\ $$

Answered by Rasheed.Sindhi last updated on 11/Oct/21

 Q=(((x+2)^4 −x^4 ))^(1/3)       =((x^4 +8x^3 +24x^2 +32x+16−x^4 ))^(1/3)       =((8x^3 +24x^2 +32x+16))^(1/3)       =2((x^3 +3x^2 +4x+2))^(1/3)       =2(((x^3 +3x^2 +3x+1)+(x+1)))^(1/3)       =2(((x+1)^3 +(x+1)))^(1/3)   The only solution,I think,is:            x+1=0             x=−1 ∈N  No natural x

$$\:\mathrm{Q}=\sqrt[{\mathrm{3}}]{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{4}} −\mathrm{x}^{\mathrm{4}} } \\ $$$$\:\:\:\:=\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{4}} +\mathrm{8x}^{\mathrm{3}} +\mathrm{24x}^{\mathrm{2}} +\mathrm{32x}+\mathrm{16}−\mathrm{x}^{\mathrm{4}} } \\ $$$$\:\:\:\:=\sqrt[{\mathrm{3}}]{\mathrm{8x}^{\mathrm{3}} +\mathrm{24x}^{\mathrm{2}} +\mathrm{32x}+\mathrm{16}} \\ $$$$\:\:\:\:=\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{3}} +\mathrm{3x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{2}} \\ $$$$\:\:\:\:=\mathrm{2}\sqrt[{\mathrm{3}}]{\left(\mathrm{x}^{\mathrm{3}} +\mathrm{3x}^{\mathrm{2}} +\mathrm{3x}+\mathrm{1}\right)+\left(\mathrm{x}+\mathrm{1}\right)} \\ $$$$\:\:\:\:=\mathrm{2}\sqrt[{\mathrm{3}}]{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} +\left(\mathrm{x}+\mathrm{1}\right)} \\ $$$$\mathrm{The}\:\mathrm{only}\:\mathrm{solution},\mathrm{I}\:\mathrm{think},\mathrm{is}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{x}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}=−\mathrm{1}\:\in\mathbb{N} \\ $$$$\mathrm{No}\:\mathrm{natural}\:\mathrm{x} \\ $$

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