Question

(3-в)(3+в)(9+в²)+(4+в²)²= помогите 8 клас

Answer 1

$\displaystyle\bf\\(3-b)(3+b)(9+b^{2} )+(4+b^{2} )^{2}=\\\\\\=(3^{2} -b^{2} )(9+b^{2} )+4^{2} +2\cdot4\cdot b^{2} +(b^{2} )^{2} =\\\\\\=(9-b^{2} )(9+b^{2} )+16+8b^{2} +b^{4} =\\\\\\=9^{2} -(b^{2} )^{2} +16+8b^{2}+b^{4} =\\\\\\=81-b^{4} +16+8b^{2} +b^{4} =\\\\\\=\underbrace{(b^{4} -b^{4}) }_{0}+8b^{2} +\underbrace{(81+16)}_{97}=8b^{2} +97$