Solving the Algebraic Expression and Finding the Value of k

The question asks us to evaluate a given algebraic expression and find the value of \(k\) when the expression is written in the form \(5 \times 10^k\).

The given expression is:

\[ \frac{{{{\left( {251} \right)}^3} + {{\left( {249} \right)}^3}}}{{25.1 \times 25.1 - 624.99 + 24.9 \times 24.9}} \]

Let's analyze the numerator and the denominator separately.

Analyzing the Numerator

The numerator is \( {{\left( {251} \right)}^3} + {{\left( {249} \right)}^3} \). This is in the form of the sum of cubes, \(a^3 + b^3\).

The algebraic identity for the sum of cubes is: \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\).

Here, let \(a = 251\) and \(b = 249\).

So, the numerator becomes:

\[ (251 + 249)(251^2 - 251 \times 249 + 249^2) \]

Let's calculate the sum \(a+b\):

\[ 251 + 249 = 500 \]

So the numerator is \( 500 (251^2 - 251 \times 249 + 249^2) \).

Analyzing the Denominator

The denominator is \( 25.1 \times 25.1 - 624.99 + 24.9 \times 24.9 \).

This can be written as \( (25.1)^2 - 624.99 + (24.9)^2 \).

Notice the numbers \(25.1\) and \(24.9\) are related to \(251\) and \(249\) from the numerator. Specifically, \(25.1 = \frac{251}{10}\) and \(24.9 = \frac{249}{10}\).

Let's examine the middle term \(624.99\). Let's try multiplying \(25.1\) and \(24.9\):

\[ 25.1 \times 24.9 = (25 + 0.1) \times (25 - 0.1) \]

Using the identity \((x+y)(x-y) = x^2 - y^2\), with \(x=25\) and \(y=0.1\):

\[ (25 + 0.1)(25 - 0.1) = 25^2 - (0.1)^2 = 625 - 0.01 = 624.99 \]

So, the middle term \(624.99\) is actually \(25.1 \times 24.9\).

Now, the denominator can be written as \( (25.1)^2 - (25.1 \times 24.9) + (24.9)^2 \).

Let \(x = 25.1\) and \(y = 24.9\). The denominator is \( x^2 - xy + y^2 \).

Simplifying the Expression

Now let's put the numerator and denominator back together. We can express the numerator's terms in terms of \(x\) and \(y\):

Since \(251 = 10 \times 25.1 = 10x\) and \(249 = 10 \times 24.9 = 10y\).

The numerator \( (251)^3 + (249)^3 \) is \( (10x)^3 + (10y)^3 \).

\[ (10x)^3 + (10y)^3 = 1000x^3 + 1000y^3 = 1000(x^3 + y^3) \]

Using the sum of cubes formula for \(x^3 + y^3\):

\[ 1000(x+y)(x^2 - xy + y^2) \]

The expression becomes:

\[ \frac{1000(x+y)(x^2 - xy + y^2)}{x^2 - xy + y^2} \]

Assuming \(x^2 - xy + y^2 \neq 0\), which is true for distinct non-zero \(x\) and \(y\), we can cancel the term \( (x^2 - xy + y^2) \) from the numerator and the denominator.

The simplified expression is \( 1000(x+y) \).

Calculating the Final Value

Substitute the values of \(x\) and \(y\) back into the simplified expression:

\[ x+y = 25.1 + 24.9 = 50.0 = 50 \]

The value of the expression is:

\[ 1000 \times (x+y) = 1000 \times 50 = 50000 \]

Finding the Value of k

The problem states that the value of the expression is \( 5 \times 10^k \).

We found the value to be \( 50000 \).

So, we have the equation:

\[ 5 \times 10^k = 50000 \]

To find \(10^k\), divide both sides by 5:

\[ 10^k = \frac{50000}{5} = 10000 \]

We need to express \(10000\) as a power of 10.

\[ 10000 = 10 \times 10 \times 10 \times 10 = 10^4 \]

So, \( 10^k = 10^4 \).

Comparing the exponents, we find \(k = 4\).

The value of \(k\) is 4.

Additional Information: Algebraic Identities

Algebraic identities are equations that are true for all values of the variables involved. They are useful tools for simplifying and manipulating algebraic expressions.

• Sum of Cubes: \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\). This identity was crucial in simplifying the numerator of the given expression.
• Difference of Cubes: \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\). This identity is used when dealing with the difference of two cubed terms.
• Difference of Squares: \(a^2 - b^2 = (a-b)(a+b)\). We used a variation of this identity \((x+y)(x-y) = x^2 - y^2\) to confirm that \(25.1 \times 24.9 = 624.99\).
• Perfect Squares: \((a+b)^2 = a^2 + 2ab + b^2\) and \((a-b)^2 = a^2 - 2ab + b^2\). These are fundamental identities for expanding squared binomials.

Recognizing these patterns in complex expressions allows for significant simplification, making calculations much easier.