Use polynomial identities to factor the polynomials or simplify the expressions.
We are asked to simplify two expressions using algebraic identities. The expressions are \(10^3 + 5^3\) and \(8^1 - 2^1\). These expressions can be evaluated directly or by applying specific algebraic properties. We will use the sum of cubes identity, \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\), for the first expression. For the second expression, we recognize that it is a basic subtraction, but we can also relate it to power laws or properties of numbers.
The first expression is \(10^3 + 5^3\). Let \(a = 10\) and \(b = 5\). According to the sum of cubes identity, we can write:
\[10^3 + 5^3 = (10 + 5)(10^2 - 10 \times 5 + 5^2)\]
Calculating the terms inside the parentheses:
\(10 + 5 = 15\)
\(10^2 = 100\)
\(10 \times 5 = 50\)
\(5^2 = 25\)
Substituting these back, we get:
\[15(100 - 50 + 25) = 15(75)\]
Multiplying \(15 \times 75\):
\(15 \times 75 = 1125\)
So, \(10^3 + 5^3 = 1125\).
The second expression is \(8^1 - 2^1\). This is straightforward arithmetic subtraction:
\[8^1 - 2^1 = 8 - 2 = 6\]
Alternatively, noting that \(8 = 2^3\), we could write the expression as \(2^3 - 2^1 = 2(2^2 - 1) = 2(4 - 1) = 2(3) = 6\). Both methods lead to the same result. The final simplified values are \(1125\) and \(6\).
Polynomial identities are essential tools in mathematics for simplifying complex calculations, such as mental arithmetic or evaluating large powers without direct multiplication. For example, to quickly find \(101^3 - 1^3\), instead of cubing \(101\), one can use the difference of cubes identity: \((101-1)(101^2 + 101 \times 1 + 1^2) = 100(10201 + 101 + 1) = 100(10303) = 1,030,300\). Mastering these identities saves time and reduces errors in complex algebraic manipulations and daily calculations.