Divide
x^3+y^3+1-3xy by (x+y+1)
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2
(x−1+y)(x
2
+1+y
2
+x+y−xy)
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)
(x−1+y)(x
2
+1+y
2
+x+y−xy)
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
hope it's helpful to you
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