Given the equation \(x^5+5(2x^3+x+2)=5x^2(x^2+2)\) has a real root in the form of \(a-\sqrt[5]{b}\) where \(a\) and \(b\) are real positive integers. Find \((a,b)\)

Yes, it is best that you remember first few rows of the Pascal triangle. They are actual the binomial expansion coefficients, which is given by \(\begin{pmatrix} n \\ r \end{pmatrix}\). For example, when \(n=5\), we have:

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TopNewest\[x^5+5(2x^3+x+2)=5x^2(x^2+2)\\ x^5+10x^3+5x+10 = 5x^4+10x^2 \\ x^5 -5x^4+10x^3-10x^2+5x+10 = 0 \\ x^5 -5x^4+10x^3-10x^2+5x-1+1 +10 = 0 \\ (x-1)^5 = -11 \\ x-1 = - \sqrt [5] {11} \\ x = 1 - \sqrt [5] {11} \\ \Rightarrow \boxed{a = 1 \quad b = 11} \]

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Oh, so we have to know the \((x-1)^5\) factorization here right?

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Yes, it is best that you remember first few rows of the Pascal triangle. They are actual the binomial expansion coefficients, which is given by \(\begin{pmatrix} n \\ r \end{pmatrix}\). For example, when \(n=5\), we have:

\[\begin{equation} \begin{split} (x-1)^5 & = \sum_{r=0}^5 {(-1)^r\begin{pmatrix} 5 \\ r \end{pmatrix}x^{5-r}} \\ & = \begin{pmatrix} 5 \\ 0 \end{pmatrix}x^5 - \begin{pmatrix} 5 \\ 1 \end{pmatrix}x^4 + \begin{pmatrix} 5 \\ 2 \end{pmatrix}x^3 - \begin{pmatrix} 5 \\ 3 \end{pmatrix}x^2 + \begin{pmatrix} 5 \\ 4 \end{pmatrix}x^1 - \begin{pmatrix} 5 \\ 5 \end{pmatrix}x^0 \\ & = x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1 \end{split} \end{equation} \]

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How you thought it will be (x-1)^5? Sir.

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Pascal triangle:

\[1 \\ 1 \quad 1 \\ 1 \quad 2 \quad 1 \\ 1 \quad 3 \quad 3 \quad 1 \\ 1 \quad 4 \quad 6 \quad 4 \quad 1 \\ 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1\\ 1 \quad 6 \quad 15 \quad 20 \quad 15 \quad 6 \quad 1 \]

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