Question

factorise -
t raised to power 6 - 7t cube -8

please help me , can't solve it....
its a ninth class question

Answer 1

$t^6 - 7 t^3 - 8 \\ \\ (t^3 - 8)(t^3 + 1) \\ \\ (t^3 - 2^3)(t^3+1^3) \\ \\ (t -2)(t^2+2t+4)(t+1)(t^2-t+1)$

for factorization some times, it is easier to find factors by looking at the coefficients. constant term divided by coefficient of highest power term.

- 8 / 1 : factors are :  +- 8 /1 , +- 4/1 , +- 2/1  +- 1/1 : 

some of these eight are possible values that appear as constants in factors. In this case the factors are : (x -8) (x +1)