Multiply: [tex]4 x \sqrt[3]{4 x^2}\left(2 \sqrt[3]{32 x^2}-x \sqrt[3]{2 x}\right)[/tex]
Answer (2)
To multiply the expression 4 x 3 4 x 2 ( 2 3 32 x 2 − x 3 2 x ) , distribute and simplify both terms to get 32 x 2 3 2 x − 8 x 3 . The final result is 32 x 2 3 2 x − 8 x 3 .
;
Distribute 4 x 3 4 x 2 across the terms in the parenthesis.
Simplify the first term: 4 x 3 4 x 2 ⋅ 2 3 32 x 2 = 32 x 2 3 2 x .
Simplify the second term: 4 x 3 4 x 2 ⋅ x 3 2 x = 8 x 3 .
Combine the simplified terms: 32 x 2 3 2 x − 8 x 3 . The final answer is 32 x 2 3 2 x − 8 x 3 .
Explanation
Understanding the Problem We are asked to multiply and simplify the expression 4 x 3 4 x 2 ( 2 3 32 x 2 − x 3 2 x ) . Let's break this down step by step to make sure we get it right!
Distributing the Term First, distribute the term 4 x 3 4 x 2 to both terms inside the parentheses:
4 x 3 4 x 2 ⋅ 2 3 32 x 2 − 4 x 3 4 x 2 ⋅ x 3 2 x
Simplifying the First Term Now, let's simplify each term separately. For the first term:
4 x 3 4 x 2 ⋅ 2 3 32 x 2 = 8 x 3 4 x 2 ⋅ 32 x 2 = 8 x 3 128 x 4
Further Simplification We can simplify 3 128 x 4 further. Since 128 = 2 7 , we can write it as 2 6 ⋅ 2 = ( 2 2 ) 3 ⋅ 2 = 4 3 ⋅ 2 . Also, x 4 = x 3 ⋅ x . So,
3 128 x 4 = 3 4 3 ⋅ 2 ⋅ x 3 ⋅ x = 4 x 3 2 x
Substituting Back Substituting this back into the first term:
8 x 3 128 x 4 = 8 x ⋅ 4 x 3 2 x = 32 x 2 3 2 x
Simplifying the Second Term Now, let's simplify the second term:
4 x 3 4 x 2 ⋅ x 3 2 x = 4 x 2 3 4 x 2 ⋅ 2 x = 4 x 2 3 8 x 3
Further Simplification Since 3 8 x 3 = 2 x , the second term becomes:
4 x 2 3 8 x 3 = 4 x 2 ⋅ 2 x = 8 x 3
Combining the Terms Now, combine the simplified terms:
32 x 2 3 2 x − 8 x 3
Final Answer So, the final simplified expression is 32 x 2 3 2 x − 8 x 3 .
Examples
Imagine you're calculating the volume of a complex 3D shape that involves cubes and irregular forms. Simplifying expressions with radicals, like in this problem, can help you break down the volume calculation into manageable parts. For instance, if you have a sculpture made of crystal with dimensions involving cube roots, this type of simplification allows you to determine the exact amount of material needed or the space it occupies. This is also useful in physics when dealing with quantities that scale with fractional exponents, such as in wave mechanics or thermodynamics.
