Question

The factors of x³-1+y³+3xyare
A. (x-1+y)(x²+1+y²+x+y-xy)
B. (x+y+1)(x²+y²+1-xy-x-y)
C. (x-1+y)(x²-1-y²+x+y+xy)
D. 3(x+y-1)(x²+y²-1)

Answer 1

The factors of the given expression $x^3-1+y^3+3xy$is $(x-1+y)(x^2+1+y^2+x+y-xy)$

Step-by-step explanation:

Given expression is $x^3-1+y^3+3xy$

To find the factors of the given expression :

• From the options we can verify the factors .
• First taking the expression $(x-1+y)(x^2+1+y^2+x+y-xy)$
• Now we can verify whether the $(x-1+y)(x^2+1+y^2+x+y-xy)$ satisfies the given expression

• Solving the expression $(x-1+y)(x^2+1+y^2+x+y-xy)$
• $(x-1+y)(x^2+1+y^2+x+y-xy)$
• $=x(x^2+1+y^2+x+y-xy)-1(x^2+1+y^2+x+y-xy)+y(x^2+1+y^2+x+y-xy)$ ( by using the distributive property )
• $=x(x^2)+x(1)+x(y^2)+x(x)+x(y)+x(-xy)-1(x^2)-1(1)-1(y^2)-1(x)-1(y)-1(-xy)+y(x^2)+y(1)+y(y^2)+y(x)+y(y)+y(-xy)$
• $=x^{1+2}+x+xy^2+x^{1+1}+xy-x^{1+1}y-x^2-1-y^2-x-y+xy+x^2y+y+y^{1+2}+xy+y^{1+1}-xy^{1+1}$ ( by using the property $a^m.a^n=a^{m+n}$ )
• $=x^3+x+xy^2+x^2+xy-x^2y-x^2-1-y^2-x-y+xy+x^2y+y+y^3+xy+y^2-xy^2$ ( adding the like terms )

• $=x^3+3xy-1+y^3$
• Rewritting we get
• $=x^3-1+y^3+3xy$
• Therefore $(x-1+y)(x^2+1+y^2+x+y-xy)=x^3-1+y^3+3xy$

Therefore option A)  $(x-1+y)(x^2+1+y^2+x+y-xy)$ is correct

The factors of the given expression $x^3-1+y^3+3xy$is $(x-1+y)(x^2+1+y^2+x+y-xy)$