Question

If `x^3 - ( x+1) ^2 = 2001` then what is the value of x?

Answer 1

`x=13" "` or `" "x = -6+-sqrt(118)i`

Explanation:

Given:

`x^3-(x+1)^2=2001`

Multiply out the left hand side to get:

`x^3-x^2-2x-1=2001`

Subtract `2001` from both sides to get:

`x^3-x^2-2x-2002 = 0`

Note that the signs of the coefficients are in the pattern `+ - - -`. With one change of signs, Descartes' Rule of Signs tells us that this cubic equation has exactly one positive real solution.

The prime factorisation of `2002` is:

`2002 = 2 * 7 * 11 * 13`

So the positive factors of `2002` are:

`1, 2, 7, 11, 13, 14, 22, 26, 77, 91, 143, 154, 182, 286, 1001, 2002`

We could try each in turn, but note that `root(3)(2002) ~~ 12.6`, so try `13` first and find:

`(color(blue)(13))^3-(color(blue)(13))^2-2(color(blue)(13))-2002 = 2197-169-26-2002 = 0`

So `x=13` is a solution and `(x-13)` a factor:

`0 = x^3-x^2-2x-2002 = (x-13)(x^2+12x+154)`

The remaining quadratic has no real zeros, as we can see from its discriminant, but we can factor it using complex numbers:

`0 = x^2+12x+154`

`color(white)(0) = x^2+12x+36+118`

`color(white)(0) = (x+6)^2+(sqrt(118))^2`

`color(white)(0) = (x+6)^2-(sqrt(118)i)^2`

`color(white)(0) = ((x+6)-sqrt(118)i)((x+6)+sqrt(118)i)`

`color(white)(0) = (x+6-sqrt(118)i)(x+6+sqrt(118)i)`

Hence roots `x = -6+-sqrt(118)i`

Answer 2

`x=13`

Explanation:

Here,

`x^3-(x+1)^2=2001`

`=>x^3-x^2-2x-1=2001`

`=>x^3-x^2-2x-2002=0`

`=>x^3-2197-x^2+169-2x+26=0`

`=>(x^3-13^3)-(x^2-13^2)-2(x-13)=0`

`=>(x-13)(x^2+13x+169)-(x-13)(x+13)`

`color(white)(..........................................................)-2(x-13)=0`

`=>(x-13)[x^2+13x+169-x-13-2]=0`

`=>(x-13)[x^2+12x+154]=0`

`=>x-13=0 or x^2+12x+154=0`

`(i)x-13=0=>x=13`

`(ii)x^2+12x+154=0=>a=1 ,b=12 ,c=154`

`triangle=b^2-4ac=144-4(1)(154)=144-616`

`=>triangle=-472 <0=>x!inRR` ,but `x in CC`

Hence, `x=13`

Note:
`x^3-(x+1)^2=2197-196`
`x^3-(x+1)^2=(13)^3-(14)^2`
`x^3-(x+1)^2=(13)^3-(13+1)^2=>x=13`

For `x inCC` please see answer below given by @George C.