Answer
$t^{24}-4t^{18}+6t^{12}-4t^6+1$
Work Step by Step
RECALL:
$x^my^m=(xy)^m$
Use the rule above to obtain:
$=[(t^3-1)(t^3+1)]^4
\\=(t^3-1)(t^3+1) \cdot (t^3-1)(t^3+1) \cdot (t^3-1)(t^3+1) \cdot (t^3-1)(t^3+1)$
Use the formula $(a-b)(a+b) = a^2-b^2$ with $a=t^3$ and $b=1$ to obtain:
$=[(t^3)^2-1^2] \cdot [(t^3)^2-1^2] \cdot [(t^3)^2-1^2] \cdot [(t^3)^2-1^2]
\\=(t^6-1) \cdot (t^6-1) \cdot (t^6-1) \cdot (t^6-1)
\\=[(t^6-1) \cdot (t^6-1)] \cdot [(t^6-1) \cdot (t^6-1)]$
Use the formula $(a-b)(a-b) = a^2-2ab+b^2$ with $a=t^6$ and $b=1$ to obtain:
$=[(t^6)^2-2(t^6)(1)+1^2] \cdot [(t^6)^2-2(t^6)(1)+1^2]
\\=(t^{12}-2t^6+1) \cdot (t^{12}-2t^6+1)$
Distribute each term of the first trinomial to the second trinomial to obtain:
$=t^{12}(t^{12}-2t^6+1)-2t^6(t^{12}-2t^6+1)+1(t^{12}-2t^6+1)
\\=t^{12}(t^{12}) -t^{12}2t^6+t^{12}(1)-2t^6(t^{12})-2t^6(-2t^6)-2t^6(1)+(t^{12}-2t^6+1)
\\=t^{24}-2t^{18}+t^{12}-2t^{18}+4t^{12}-2t^6+t^{12}-2t^6+1$
Combine like terms to obtain:
$\\=t^{24}+(-2t^{18}-2t^{18})+(t^{12}+4t^{12}+t^{12})+(-2t^6-2t^6)+1
\\=t^{24}+(-4t^{18})+6t^{12}+(-4t^6)+1
\\=t^{24}-4t^{18}+6t^{12}-4t^6+1$
