factorise x³+y³-1+3xy
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x 3 −1+y 3 +3xy =(x) 2 +(−1) 3 +y 3 −3×x×(−1)×y
We know, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −ab+bc−ca)
x 3 −1+y 3 +3xy =[x+(−1)+y][(x) 2 +(−1) 2 +(y) 2 −(x)(−1)−(−1)(y)−(y)(x)] =(x−1+y)(x 2 +1+y 2 +x+y−xy)
Step-by-step explanation:
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