I can't solve this algebraic sum and I need some help please.

[winwin.69]

Stem: x + 1/x = a

Question: Prove that:

(x³ + 1/x³)(x² + 1/x²) = a⁵ - 5a³ + 6a

 

[MarkFL]

Hello, and welcome to FMH!

I am assuming that given:

[MATH]x+\frac{1}{x}=a[/MATH]
Prove:

[MATH]\left(x^3+\frac{1}{x^3}\right)\left(x^2+\frac{1}{x^2}\right)=a^5-5a^3+6a[/MATH]
Let's square the given to get:

[MATH]x^2+2+\frac{1}{x^2}=a^2\implies x^2+\frac{1}{x^2}=a^2-2[/MATH]
What do you find when you cube the given?

 

[MarkFL]

To follow up, when cubing the given, we find:

[MATH]x^3+3x+3\frac{1}{x}+\frac{1}{x^3}=a^3[/MATH]
[MATH]x^3+3a+\frac{1}{x^3}=a^3[/MATH]
[MATH]x^3+\frac{1}{x^3}=a^3-3a[/MATH]
And so:

[MATH]\left(x^3+\frac{1}{x^3}\right)\left(x^2+\frac{1}{x^2}\right)=(a^3-3a)(a^2-2)=a^5-5a^3+6a[/MATH]
Shown as desired.

 

[Deleted member 4993]

Another equation that could have been used:

p³ + q³ = (p + q) (p² - pq + q²)

x³ + 1/x³ = (x + 1/x) * (x² - 1 + 1/x²)

 

[Sonal7]

I am struggling with the same type of problem. I am posting as a new thread.

 

[Steven G]

OK.