Question

Factorize x⁴-14x²y²+y⁴

Answer 1

Answer:

x^4 +(x^2*y^2 )+y^4 + (x^2*y^2 )-(x^2*y^2)

=x^4+ 2(x^2*y^2) + y^4 - (x^2*y^2)

= {(X^2)^2 + 2 (x^2*y^2) + (y^2)^2 }- (x^2* y^2)

The above statement is in the form (a+ b)^2= a^2 + 2a*b +b^2

= {(x^2 + y^2)^2 }- (x^2 + y^2)

Taking (x^2 + y^2) common

= (x^2 + y^2)(x^2 +y^2 -1)

Answer 2

Answer:

Step-by-step explanation:

This can  be factored by completing the square. 

Recall that perfect square trinomial  x²+2xy+y² = (x+y)².

x⁴ + x²y² +y⁴ first term and 3rd term  are perfect squares

= (x²)² +  x²y² + (y²)²

the middle term should to be 2x²y²  to make                                        it a perfect square, so add x²y² - x²y² 

= (x²)² +  x²y² + (y²)²  + x²y² - x²y²   combine 2nd and 4th terms

= (x²)² +  2 x²y² + (y²)²  - x²y²      first 3 terms form a perfect square trionomial

= (x² + y²)²- (xy)²    

it is a difference of two squares, now factor

=(x² + y² + xy)(x² + y² - xy)