Answer:
t=64
Step-by-step explanation:
L.C.M. of 2 and 3=6
[tex]t^{\frac{1}{6} } =x\\squaring\\ t^{\frac{1}{3} } =x^{2} \\cubing\\t^{\frac{1}{2} } =x^{3} \\x^{2} +x^{3} =12\\when~ x=2\\2^2+2^3=4+8=12\\so~ x=2~ is~ one~ root.\\[/tex]
By synthetic division.
2| 1 1 0 -12
| 2 6 12
-------------------
1 3 6|0
x²+3x+6=0
disc.=b²-4ac=3²-4×1×6=9-24=-15<0
so roots are imaginary.
x=2
[tex]t^{\frac{1}{6} } =2\\t=2^6=64[/tex]
if the statement is as given by Sadievigil then he or she is correct otherwise refer to my solution.
