Question

2. Factorise 27d? + b3 + 8c3 - 18abc using identity.​

Answer 1

${27a}^{3} + {b}^{3} + {8c}^{3} - 18abc \\ \\ {(3a)}^{3} + {(b)}^{3} + {(2c)}^{3} - 3(3a)(b)(2c) \\ \\ (3a + b + 2c)( {9a}^{2} + {b}^{2} + {4c}^{2} - 3ab - 2bc - 6ac )$

$<marquee>MARK BRAINLIEST✌✌✌$