MHB Efficient Method for Extracting Square Root of Complex Expressions

[NotaMathPerson]

Hello!

Is there a way to extract the square root of this expression without expanding? Please teach me how to go about it.

$4\left((a^2-b^2)cd+ab(c^2-b^2)\right)^2+\left((a^2-b^2)(c^2-b^2)-4abcd\right)^2$

I tried expanding it and it was very laborious and I end up not getting the correct answer.

 

[suluclac]

Hello,
I don't see a square root in the expression. Are you asking us how to factor the expression?

 

[NotaMathPerson]

Hello,
I don't see a square root in the expression. Are you asking us how to factor the expression?

Hello! This problem is from a book and it says that I have to extract the square root of the expression.

 

[suluclac]

Does the problem from the book say
4((a² - b²)cd + ab(c² - b²))² + ((a² - b²)(c² - b²) - 4abcd)²?

 

[NotaMathPerson]

Does the problem from the book say
4((a² - b²)cd + ab(c² - b²))² = ((a² - b²)(c² - b²) - 4abcd)²?

Here's the screen shot from the book.

 

[suluclac]

I'll take that as a no.

 

[NotaMathPerson]

Hello!
I just finished solving the problem!

Here is how I solved it

I let $(a^2-b^2)=x$ and $(c^2-d^2)=y$

Now we have

$4(xcd+yab)^2+(xy-4abcd)^2$

expanding the terms

$4x^2c^2d^2+8xyabcd+4y^2a^2b^2+x^2y^2-8xyabcd+16a^2b^2c^2d^2$

Simplifying

$4x^2c^2d^2+4y^2a^2b^2+x^2y^2+16a^2b^2c^2d^2$

By using factoring

$x^2(4c^2d^2+y^2)+4a^2b^2(y^2+4c^2d^2) = (x^2+4a^2b^2)(4c^2d^2+y^2)$

Substituting the value of x and y$\left((a^2-b^2)^2+4a^2b^2\right) \left((c^2-d^2)^2+4c^2d^2\right)$

By expanding and some simplifications

$(a^4+2a^2b^2+b^4)(c^4+2c^2d^2+d^4)$Both factors are square of binomials

$(a^2+b^2)^2(c^2+d^2)^2$

Taking the square root

$(a^2+b^2)(c^2+d^2)$

I guess expansion is really necessary in this problem.

 

[suluclac]

Correct.