Решите пожалуйста номер 847

Решите задачу:

$p^2-y^2=(p-y)(p+y)\\\\k^2-c^2=(k-c)(k+c)\\\\1-p^2=(1-p)(1+p)\\\\0,25a^4-0,49=(0,5a^2-0,7)(0,5a^2+0,7)\\\\16-b^2=(4-b)(4+b)\\\\25a^2-9b^2=(5a-3b)(5a+3b)\\\\a^2-9=(a-3)(a+3)\\\\a^4-b^4=(a^2-b^2)(a^2+b^2)=(a-b)(a+b)(a^2+b^2)$

$x^6-y^8=(x^3)^2-(y^4)^2=(x^3-y^4)(x^3+y^4)\\\\p^4-0,16=(p^2)^2-(0,4)^2=(p^2-0,4)(p^2+0,4)=\\\\=(p-\frac{\sqrt{10}}{5})(p+\frac{\sqrt{10}}{5})(p^2+0,4)\\\\\star\; \; 0,4=(\sqrt{0,4})^2=\left (\sqrt{\frac{4}{10}}\right )^2=\left (\frac{2}{\sqrt{2\cdot 5}}\right )^2=\left(\frac{\sqrt2}{\sqrt5}\right )^2=\left (\frac{\sqrt{10}}{5}\right )^2\\\\a^2b^2-36x^4=(ab)^2-(6x^2)^2=(ab-6x^2)(ab+6x^2)=\\\\=(\sqrt{ab}-\sqrt6x)(\sqrt{ab}+\sqrt6x)(ab+6x^2)$