Question
Show the following quadratic equation by factorization method:
$x^2 + 10ix - 21 = 0$
$x^2 + 10ix - 21 = 0$
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$(x+1)^4-4(x+1)^3(x-1)+6(x+1)^2(x-1)^2-4(x+1)(x-1)^3+(x-1)^4$
$(a+b)+\left(a \omega+b \omega^2\right)+\left(a \omega^2+b \omega\right)=0$