Question

2 x cube minus 5 y square bracket close into bracket open 2 by 3 + 3 y square bracket close​

Answer 1

❤❤PLEASE FIND UR ANSWER❤❤
${eq}^{n \: } is \: ({2x}^{3} - {5y}^{2} ) \times (2 \times 3 + {3y}^{2} ) \\ \\ ({2x}^{3} - {5y}^{2} ) \times (6 + {3y}^{2} ) \\ {2x}^{3 } \times 6 + {2x}^{3} \times {3y}^{2} - {5y }^{2} \times 6 - {5y}^{2} \times {3y}^{2} \\ {12x}^{3} + 6 {y}^{2} {x}^{3} - {30y}^{2} - {15y}^{4} \\ \\ since \: now \: this \: e {q}^{n \: } cannot \: be \: solved \: \\ further \: so \: ur \: ans \: is \\ \\ {12x}^{3} + {6y}^{2} {x}^{3} - {30y}^{2} - {15y}^{4 } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: or \\ {6x}^{3} (2 + {y}^{2} ) - {15y}^{2} (2 + {y}^{2} ) \\ ie \\ (2 + {y}^{2} )( {6x}^{3} - 15 {y}^{2} )$
✌✌HOPE IT HELPS✌✌

Answer 2

ur answer is

(2+y^2)(6x^3-15y^2)