Question

Factorise: a² + 4b2 + c2/9+ 4ab -4bc/3 - 2ca/3​

Answer 1

Step-by-step explanation:

${a}^{2} + 4 {b}^{2} + \frac{ {c}^{2} }{9} + 4ab - \frac{4bc}{3} - \frac{2ca}{3} \\ = {a}^{2} + {(2b)}^{2} + {( \frac{c}{3} )}^{2} + 2(2ab) - 2( \frac{2bc}{3} ) - 2( \frac{ca}{3} ) \\ = {(a + 2b - \frac{c}{3} )}^{2} \\ = (a + 2b - \frac{c}{3} ) (a + 2b - \frac{c}{3} )$