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Решите задачу:

$27k^6+64t^3=(3k+4t)(9k^2-24kt+16t^2)$
$k^4-r^6=(k^2-r^3)(k^2+r^3)$
$8w^6-s^9=(2w^2-s^3)(4w^4+4w^2s^3+s^6)$
$n^9+u^3=(n^3+u)(n^6-2n^3u+u^2)$
$8t^3+27r^3=(2t+3r)(4t^2-12tr+9r^2)$
$y^6+s^9=(y^2+s^3)(y^4-2y^2s^3+s^6)$
$9p^8-81f^6=$ $(3p^4-9f^3)(3p^4+9f^3)$
$27m^9-t^3=(3m^3-t)(6m^6+54m^3t++t^6)$
$y^6-2y^3u^4+u^8=$
$t^2+8tr^2+16r^4=(t+4r^2)^2$
$u^6+2u^3p^2+p^4=(u^3+p^2)^2$
$16p^8-k^2=(4p^4-k)(4p^4+k)$
$81m^6z-49zr^2=z(81m^6-49r^2)=z(9m^3+7r)(9m^3-7r)$

$25t^6-10t^3r+r^2=(5t^3-r)^2$
$y^6-s^9=(y^2-s^3)(y^4+2y^2y^3+s^6)$
$n^4+14n^2x^2+49x^4=(n^2+7x^2)^2$

Разложите ** множителиии! СРОЧНО!