tactorize \( m^{3}+n^{3} \) \( a^{1 x}-b^{6} \) \( 8 x^{3}-y^{9} \) \( x^{3}+8 y^{3} \)

To factor \( m^{3}+n^{3} \), you can use the sum of cubes formula: \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \). So, it factors to \( (m + n)(m^2 - mn + n^2) \). For \( a^{1 x} - b^{6} \), if we assume it's \( a^x - b^6\), the expression can be factored as a difference of two squares if we treat it like a polynomial. However, with \( a \) and \( b \) being unknowns, the simpler representation remains \( a^x - b^6 \). To tackle \( 8x^{3}-y^{9} \), we can factor this as follows: recognize \( 8x^{3} \) as \( (2x)^{3} \) and \( y^{9} \) as \( (y^{3})^{3} \), allowing us to apply the difference of cubes formula: \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). Therefore, it factors to \( (2x - y^3)((2x)^2 + (2x)y^3 + (y^3)^2) \). For \( x^{3}+8y^{3} \), likewise, recognize \( 8y^{3} \) as \( (2y)^{3} \) and apply the sum of cubes formula. Thus, it factors to \( (x + 2y)(x^2 - 2xy + 4y^2) \). Happy factoring!