Question

Найдите значение выражения (3x²+2y³)(2y³-3x²) при x⁴=⅓, y²=3

Answer 1

$(3x^{2} +2y^{3})(2y^{3}-3x^{2})=-(3x^{2}+2y^{3})(3x^{2}-2y^{3})=-[(3x^{2})^{2}-(2y^{3})^{2}]=-(9x^{4}-4y^{6})\\\\x^{4}=\frac{1}{3}\\\\y^{2}=3\Rightarrow y^{6}=(y^{2})^{3}=3^{3}=27\\\\\\-(9x^{4}-4y^{6})=-(9*\frac{1}{3}-4*27)=-(3-108)=-(-105)=105$