-1
$\begingroup$

Good day, i have this exercise:

$(t + k)^{3} - t - k$ , and i must factor it.

I try this:

$(t + k)(t^2 -tk + k^2) - (t + k)$

but obviously it is not yet fully factored. The parenthesis that is trinomial, looks like a binomial square, but it is not, since the second term is not $(2 * t * k)$, since if it were, $t$ or $k$ should be $1/2$ and therefore in its roots it would come out $1/4$

I tried to look for help in symbolab, but says it can not be factored anymore.

so, how i can factor it? , The final factoring should be:

$(t + k)(t + k + 1)(t +k -1)$

$\endgroup$
3
  • $\begingroup$ I try this You got the sign wrong there, $\,-t-k=-(t\color{red}{+}k)\,$. $\endgroup$
    – dxiv
    Mar 5, 2018 at 22:39
  • $\begingroup$ yes, now i have two common terms, but I do not know how to continue $\endgroup$
    – ESCM
    Mar 5, 2018 at 23:14
  • $\begingroup$ So you have $\,a^3 - a\,$ where $\,a=t+k\,$. Factor out the common term in $\,a^3-a\,$ and see what's left. $\endgroup$
    – dxiv
    Mar 5, 2018 at 23:23

1 Answer 1

Reset to default
1
$\begingroup$

Hint: $$-t-k=-(t+k) $$

Then, use $$a^2-b^2 = (a-b)(a+b)$$

${{{{{{{}}}}}}} $

$\endgroup$
7
  • $\begingroup$ why? how ? i cant understand that $\endgroup$
    – ESCM
    Mar 5, 2018 at 23:12
  • $\begingroup$ $(t+k)^3 - t - k = (t+k)^3 - (t+k) $. Can you see the common factor here? $\endgroup$
    – krirkrirk
    Mar 5, 2018 at 23:17
  • $\begingroup$ Isn't a difference of square or difference of cubes, i can't see the common factor, help me please $\endgroup$
    – ESCM
    Mar 5, 2018 at 23:19
  • $\begingroup$ Before using the second part of my answer you need to factor $(t+k)^3-(t+k) $. I'm sure you can do it if you give it a try. Hint : how would you factor $x^3-x $? $\endgroup$
    – krirkrirk
    Mar 5, 2018 at 23:23
  • 1
    $\begingroup$ thanks, finally: $(t+k)((t+k)^{2} + 1)$ that can be: $(t + k)(t + k -1)(t+k+1)$, thanks for square difference !! and the $x^3 - x$ $\endgroup$
    – ESCM
    Mar 5, 2018 at 23:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .