$\sqrt{10^{6}\left(10^{6}+1\right)\left(10^{6}+2\right)\left(10^{6}+3\right)+1}$

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Algebra 2

Simplify the product of two radical expressions having same variable

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\sqrt{10^{6}\left(10^{6}+1\right)\left(10^{6}+2\right)\left(10^{6}+3\right)+1}

Given Expression Analysis: We are given the expression $\sqrt{(10^{6}(10^{6}+1)(10^{6}+2)(10^{6}+3)+1)}$ . Notice that the expression inside the square root is very close to the form of a squared term, specifically $(a^2 + b^2)$ where $a$ is the product of the four terms and $b$ is $1$ . We will check if this expression can be rewritten as a perfect square.
Denoting Terms: Let's denote $10^{6}$ as ' $a$ ' for simplicity. Then the expression inside the square root becomes $(a(a+1)(a+2)(a+3)+1)$ . We want to see if this can be expressed as $(\text{something})^2$ .
Factoring Attempt: We can try to factor the expression in the form of $a^2 + b^2$ by looking for a pattern. The expression $a(a+1)(a+2)(a+3)+1$ resembles the expansion of $a^2 + 1$ ^ $2$ = a^ $4$ + $2$ a^ $2$ + $1$ , but with extra terms involving 'a'. We need to check if the given expression can be factored into a perfect square.
Comparison with Perfect Squares: Let's expand $(a^2 + 1)^2$ to see if it matches our expression: $\newline$ $(a^2 + 1)^2 = a^4 + 2a^2 + 1$ $\newline$ Now, let's compare this with our expression $a(a+1)(a+2)(a+3)+1$ : $\newline$ $a(a+1)(a+2)(a+3) = a^4 + 6a^3 + 11a^2 + 6a$ $\newline$ Adding $1$ to both gives us $a^4 + 6a^3 + 11a^2 + 6a + 1$ .
Adjusting for Perfect Square: We can see that the expression $a^4 + 6a^3 + 11a^2 + 6a + 1$ is not a perfect square because it does not match the expansion of $(a^2 + 1)^2$ . However, we can notice that the expression is very close to $(a^2 + a + 1)^2$ , which expands to $a^4 + 2a^3 + 3a^2 + 2a + 1$ . Let's compare this with our expression.
Adding and Subtracting Terms: Expanding $(a^2 + a + 1)^2$ gives us: $\newline$ $(a^2 + a + 1)^2 = a^4 + 2a^3 + 3a^2 + 2a + 1$ $\newline$ Comparing this with our expression $a^4 + 6a^3 + 11a^2 + 6a + 1$ , we see that the coefficients of $a^3$ and $a^2$ are larger in our expression, and the coefficient of $a$ is smaller.
Grouping for Perfect Square: We need to adjust the terms to make our expression a perfect square. Let's try adding and subtracting the same value inside the square root to create a perfect square without changing the value of the expression. We can add and subtract $(3a^3 + 3a^2 + a)$ , which is the difference between our expression and $(a^2 + a + 1)^2$ .
Taking Square Root: Adding and subtracting $(3a^3 + 3a^2 + a)$ inside the square root gives us: $\newline$ $\sqrt{a^4 + 6a^3 + 11a^2 + 6a + 1 + 3a^3 + 3a^2 + a - 3a^3 - 3a^2 - a}$ $\newline$ This simplifies to: $\newline$ $\sqrt{(a^4 + 6a^3 + 11a^2 + 6a + 1) + (3a^3 + 3a^2 + a) - (3a^3 + 3a^2 + a)}$
Substitution and Final Expression: Now, we can group the terms to form a perfect square: $\newline$ $\sqrt{(a^4 + 6a^3 + 11a^2 + 6a + 1) + (3a^3 + 3a^2 + a) - (3a^3 + 3a^2 + a)}$ $\newline$ = $\sqrt{(a^4 + 6a^3 + 11a^2 + 6a + 1) + (3a^3 + 3a^2 + a) + 1 - (3a^3 + 3a^2 + a) - 1}$ $\newline$ = $\sqrt{(a^2 + a + 1)^2}$
Substitution and Final Expression: Now, we can group the terms to form a perfect square: $\newline$ $\sqrt{(a^4 + 6a^3 + 11a^2 + 6a + 1) + (3a^3 + 3a^2 + a) - (3a^3 + 3a^2 + a)}$ $\newline$ = $\sqrt{(a^4 + 6a^3 + 11a^2 + 6a + 1) + (3a^3 + 3a^2 + a) + 1 - (3a^3 + 3a^2 + a) - 1}$ $\newline$ = $\sqrt{(a^2 + a + 1)^2}$ Since we have a perfect square under the square root, we can take the square root of the expression: $\newline$ $\sqrt{(a^2 + a + 1)^2} = a^2 + a + 1$
Substitution and Final Expression: Now, we can group the terms to form a perfect square: $\newline$ $\sqrt{(a^4 + 6a^3 + 11a^2 + 6a + 1) + (3a^3 + 3a^2 + a) - (3a^3 + 3a^2 + a)}$ $\newline$ = $\sqrt{(a^4 + 6a^3 + 11a^2 + 6a + 1) + (3a^3 + 3a^2 + a) + 1 - (3a^3 + 3a^2 + a) - 1}$ $\newline$ = $\sqrt{(a^2 + a + 1)^2}$ Since we have a perfect square under the square root, we can take the square root of the expression: $\newline$ $\sqrt{(a^2 + a + 1)^2} = a^2 + a + 1$ Now, we substitute back the value of 'a' which is $10^{6}$ : $\newline$ $a^2 + a + 1 = (10^{6})^2 + 10^{6} + 1$ $\newline$ = $10^{12} + 10^{6} + 1$
Substitution and Final Expression: Now, we can group the terms to form a perfect square: $\newline$ $\sqrt{(a^4 + 6a^3 + 11a^2 + 6a + 1) + (3a^3 + 3a^2 + a) - (3a^3 + 3a^2 + a)}$ $\newline$ = $\sqrt{(a^4 + 6a^3 + 11a^2 + 6a + 1) + (3a^3 + 3a^2 + a) + 1 - (3a^3 + 3a^2 + a) - 1}$ $\newline$ = $\sqrt{(a^2 + a + 1)^2}$ Since we have a perfect square under the square root, we can take the square root of the expression: $\newline$ $\sqrt{(a^2 + a + 1)^2} = a^2 + a + 1$ Now, we substitute back the value of 'a' which is $10^{6}$ : $\newline$ $a^2 + a + 1 = (10^{6})^2 + 10^{6} + 1$ $\newline$ = $10^{12} + 10^{6} + 1$ The final expression is $10^{12} + 10^{6} + 1$ , which is the value of the original expression under the square root.