Solve: Square Root of (405²+405²×406²+406²) - 405² Expression

Let's examine the problem:

$\sqrt{405^2+405^2\cdot406^2+406^2}-405^2=$

We'll focus on the expression under the square root. For calculation without a calculator, we naturally want to find a way to eliminate the square root. To do this, we need to transform the expression inside the root into a squared expression, which we'll try to do using the squared binomial formula in two main steps:

First - for simplicity of calculation and generality of solution let's denote:

$406=a\rightarrow405=a-1$

therefore the expression under the root is:

$405^2+405^2\cdot406^2+406^2\\ \downarrow\\ (a-1)^2+(a-1)^2a^2+a^2$

Step A.

We'll use a method called: "completing the square" In this method, we make the expression appear in the form of a squared binomial plus a correction term, meaning we "rearrange" the formula by moving terms within the formula itself, and use the standard form to get a structure similar to the squared binomial form:

$\boxed{ b^2-2bc+c^2= (b-c)^2}\\ \downarrow\\ \textcolor{blue}{b^2+c^2=(b-c)^2+2bc}$

Note that we can use this method on the expression in question (under the root), if we now denote:

$\textcolor{red}{ a-1=b\\ a=c }$

we get:

$(a-1)^2+(a-1)^2a^2+a^2 \\ \downarrow\\ b^2+b^2c^2+c^2\\ b^2+c^2+b^2c^2$

We'll use the "rearranged" formula we got earlier (highlighted in blue) and substitute it in the last expression we got:

$405^2+405^2\cdot406^2+406^2\\ \downarrow\\ (a-1)^2+a^2(a-1)^2+a^2 \\ \downarrow\\ b^2+b^2c^2+c^2\\ \textcolor{blue}{b^2+c^2}+b^2c^2 \\ \downarrow\\ \textcolor{blue}{(b-c)^2+2bc}+b^2c^2$

Now we'll return to parameter a using the definitions for b and c that we defined earlier (in red):

$405^2+405^2\cdot406^2+406^2\\ \downarrow\\ (a-1)^2+a^2(a-1)^2+a^2 \\ \downarrow\\ b^2+b^2c^2+c^2\\ \textcolor{blue}{b^2+c^2}+b^2c^2 \\ \downarrow\\ \textcolor{blue}{(b-c)^2+2bc}+b^2c^2 \\ \downarrow\\ ((a-1)-a)^2+2(a-1)a+(a-1)^2a^2$

where the first steps are just a reminder of what we've done so far and only the final step is substituting the expressions and returning to parameter a,

Let's continue and simplify the first term on the left in the expression we got:

$\textcolor{green}{ ((a-1)-a)^2}+2(a-1)a+(a-1)^2a^2 \\ \downarrow\\ \textcolor{green}{1}+2(a-1)a+(a-1)^2a^2$

Step B.

Now we'll use the laws of exponents:

$\boxed{x^ny^n=(xy)^2}$

and we'll express the third term from the left as a power:

$1+2(a-1)a+\textcolor{red}{(a-1)^2a^2} \\ 1+2(a-1)a+\textcolor{red}{((a-1)a)^2}$

Now let's rearrange, and factor using the shortened multiplication formula for squared binomial (in its addition form):

$1+2(a-1)a+((a-1)a)^2 \\ \downarrow\\ (\textcolor{orange}{(a-1)a})^2+2\textcolor{orange}{(a-1)a}\cdot\textcolor{purple}{1}+\textcolor{purple}{1}^2\\ \downarrow\\ (\textcolor{orange}{(a-1)a}+\textcolor{purple}{1})^2$

In the next step the square root will cancel out the squared power, which was actually the whole purpose of this development (let's summarize the solution steps):

$\sqrt{405^2+405^2\cdot406^2+406^2}-405^2= \\ \downarrow\\ \sqrt{(a-1)^2+(a-1)^2a^2+a^2 }-(a-1)^2= \\ \downarrow\\ \sqrt{\textcolor{red}{(a-1)^2+a^2}+(a-1)^2a^2 }-(a-1)^2= \\ \downarrow\\ \sqrt{\textcolor{red}{((a-1)-a)^2+2(a-1)a}+(a-1)^2a^2}-(a-1)^2\\ \downarrow\\ \sqrt{(\textcolor{orange}{(a-1)a})^2+2\textcolor{orange}{(a-1)a}\cdot\textcolor{purple}{1}+\textcolor{purple}{1}^2}-(a-1)^2\\ \downarrow\\ \sqrt{(\textcolor{orange}{(a-1)a}+\textcolor{purple}{1})^2}-(a-1)^2\\ \downarrow\\ \sqrt{((a-1)a+1)^2}-(a-1)^2= \\ \downarrow\\ (a-1)a+1-(a-1)^2$

We achieved our goal - we eliminated the root

Completing the solution:

Now we just need to expand the parentheses using the shortened multiplication formula for squared binomial and the distributive law, and then simplify the resulting expression:

$(a-1)a+1-(a-1)^2\\ \downarrow\\ a^2-a+1-(a^2-2a+1)\\ a^2-a+1-a^2+2a-1=\\ a$

Therefore we got that the result of simplifying the expression is simply:

$\sqrt{405^2+405^2\cdot406^2+406^2}-405^2= \\ \downarrow\\ \sqrt{(a-1)^2+a^2(a-1)^2+a^2 }-(a-1)^2= \\ a=406$

Therefore the correct answer is answer B

Note and additional question:

Note that the calculation was done in general form for parameter a, so now let's ask,

What will be the result of simplifying the expression?

$\sqrt{1414^2+1414^2\cdot1415^2+1415^2}-1414^2=?$