Question

factorise 3x2+27y2+z2-18xy+6root 3yz-2root3zx​

Answer 1

Solution :

/* We know the algebraic identity:

a²+b²+c²+2ab+2bc+2ca

= (a+b+c)² */

Here ,

-3x²+27y²+z²-18xy+6√3yz-2√3zx

= (-√3x)²+(3√3y)²+z²+2*(-√3x)(3√3y)+2*(3√3y)z+2*z*(-√3x)

= (-√3x+3√3y+z)²

Therefore,

-3x²+27y²+z²-18xy+6√3yz-2√3zx

= (-√3x+3√3y+z)²

= (-√3x+3√3y+z)(-√3x+3√3y+z)

•••••

Answer 2

Answer:

-3x²+27y²+z²-18xy+6√3yz-2√3zx

= (-√3x)²+(3√3y)²+z²+2*(-√3x)(3√3y)+2*(3√3y)z+2*z*(-√3x)

= (-√3x+3√3y+z)²

-3x²+27y²+z²-18xy+6√3yz-2√3zx

= (-√3x+3√3y+z)²

= (-√3x+3√3y+z)(-√3x+3√3y+z)