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If $1+9r^2 + 81r^4= 256$ and $1 + 3r + 9r^2 = 32$, then find the value of $1-3r +9r^2.$ |
8 4 16 12 |
8 |
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 |
