Quadratics using Pascal's triangle

[HorseRidingTic]

Homework Statement

The problem equation is contained in the picture.

Homework Equations

Pascal's Triangle is useful is this one.

The Attempt at a Solution

The difficulty I'm having is in going between lines 2 and 3 which I've marked with a little red dot.

The closest I get to simplifying it is = a⁴ + B⁴ + 4aB(a²+B²) + 6a²B² . From there I can't figure out the way in which to reduce it further.

P.S I also used the quadratic formula to solve this one (the one with the b² - 4ac) and my answer came to 748.52 but not quite 752. Why does the quadratic formula not work here?

Thank you for your help,
Ben

 

[Buffu]

$\alpha^4 + \beta^4 + 4\alpha^3\beta + 6\alpha^2\beta^2 + 4\alpha \beta^3 = \alpha^4 + \beta^4 + 4\alpha^3\beta + \color{red}{8}\alpha^2\beta^2 + 4\alpha \beta^3 - \color{blue}{2\alpha^2 \beta^2} = \alpha^4 + \beta^4 + 4\alpha\beta(\alpha^2 + 2\alpha\beta + \beta^2) - {2\alpha^2 \beta^2}$

Now use $a^2 + 2ab + b^2 = (a+b)^2$

 

[HorseRidingTic]

Amazing! Thank you so much Buffu :)

 

[Buffu]

 

[Mark44]

I have a minor gripe with the author of this problem.

A quadratic equation with roots $\alpha$ and $\beta$ is $(x - \alpha)(x - \beta)$, and so ...

What is shown is not an equation, since the symbol = is not present.

Again, my gripe is with the author, not the person who started this thread.

 

[Buffu]

Number of people confusing between expression and equation is surprisingly high.