can someone answer this ​

Answer: The equation is not an identity

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Explanation:

We'll expand out the right hand side to see if we can simplify it to the left hand side. The stuff on the left will stay the same for each step.

I'll use  c = a+b as a way to help distribute.

[tex]a^4 + b^4 = (a+b)(a^3 - a^2b-ab^2 + b^3)\\\\a^4 + b^4 = c(a^3 - a^2b-ab^2 + b^3)\\\\a^4 + b^4 = a^3c - a^2bc-ab^2c + b^3c\\\\a^4 + b^4 = a^3(c) - a^2b(c)-ab^2(c) + b^3(c)\\\\a^4 + b^4 = a^3(a+b) - a^2b(a+b)-ab^2(a+b) + b^3(a+b)\\\\a^4 + b^4 = a^4+a^3b - a^3b-a^2b^2-a^2b^2-ab^3 + ab^3+b^4\\\\a^4 + b^4 = a^4+(a^3b - a^3b)+(-a^2b^2-a^2b^2)+(-ab^3 + ab^3)+b^4\\\\a^4 + b^4 = a^4-2a^2b^2+b^4\\\\[/tex]

Unfortunately, we can't go any further. We have a^4 and b^4 on the right hand side, but we also have the term -2a^2b^2 that didn't cancel out. Therefore, the original equation is not an identity.

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Here's another way you can see that we don't have an identity.

Let's say we fixed 'a' to be equal to 1.

That means the equation

[tex]a^4 + b^4 = (a+b)(a^3 - a^2b-ab^2 + b^3)[/tex]

would turn into

[tex]1^4 + b^4 = (1+b)(1^3 - 1^2b-1b^2 + b^3)[/tex]

and simplify to

[tex]1 + b^4 = (1+b)(1 - b-b^2 + b^3)[/tex]

Now replace b with x and you'll have the two equations to represent the left and right hand sides

[tex]y = 1+x^4\\y = (1+x)(1-x-x^2+x^3)[/tex]

which you can graph on the same xy axis. You should note that the two equations produce different curves. The curves would need to be identical if we wanted an identity to happen.

Side note: you can pick other values for 'a' instead of just 1.

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