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Exercise

$\left(2a+b-3c+4\right)^2\:\:\left(6a+2b-3c\right)\:\:\left(a-b\right)$

Final answer to the exercise

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Other solved exercises

$121p^{36}-\frac{25}{81}x^{2}$

$b2+10b+25$

$\left(4x^2-2x+1\right)+\left(x^2-3x+1\right)-\left(5x^2-2x-3\right)$

$x^2-10x+19=0$

$2x-4-1+x$

$\frac{x\cdot sec^2x}{tamx}$

$-114-78$
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