Đáp án:
Giải thích các bước giải:
Theo bài ra ta có :
$\left\{\begin{matrix}n+26=a^3& \\n-11=b^3& \end{matrix}\right.$
`=> a^3-b^3=(n+26)-(n-11)`
`=> a^3-b^3=37`
`<=> (a-b)(a^2+ab+b^2)=37`
Vì $\left\{\begin{matrix}a-b>0& \\a^2+ab+b^2≥ 0& \end{matrix}\right.$
`TH_1:`
$\left\{\begin{matrix}a-b=1& \\a^2+ab+b^2=37& \end{matrix}\right.$`<=>`$\left\{\begin{matrix}a=1+b& \\a^2+ab+b^2=37& \end{matrix}\right.$
`<=> (b+1)^2+(b+1).b+b^2-37=0`
`<=> 3b^2+3b-36=0`
`<=> (b-3)(b+4)=0`
`<=>`\(\left[ \begin{array}{l}b=3\\b=-4(KTM)\end{array} \right.\)
`=> a=b+1=3+1=4`
`=> n=38`
`TH_2:`
$\left\{\begin{matrix}a-b=37& \\a^2+ab+b^2=1& \end{matrix}\right.$`<=>`$\left\{\begin{matrix}a=37+b& \\a^2+ab+b^2=1& \end{matrix}\right.$
`<=> (b+37)^2+(b+37).b+b^2=1`
`<=> 3b^2+111b+1368=0`
`<=> b^2+36b+456=0`
`<=> (b+37/2)^2=-455/4` ( vô lý )
Vậy `n=38`
