If $1+9r^2 + 81r^4= 256$ and $1 + 3r + 9r^2 = 32$, then find the value of $1-3r +9r^2.$
Answer & explanation
Correct answer: option 1
If $1+9r^2 + 81r^4= 256$
$1 + 3r + 9r^2 = 32$,
then find the value of $1-3r +9r^2.$
We know that,
x4 + x2y2 + y4 = (x2 – xy + y2) (x2 + xy + y2)
then the value of $1-3r +9r^2$ = \(\frac{256}{32}\) = 8