What is the irrational cube root −24−−−−√3 rewritten as a product of an integer and another irrational cube root?(1 point)
Responses
−23–√3
negative 2 root index 3 start root 3 end root
26–√3
2 root index 3 start root 6 end root
23–√3
2 root index 3 start root 3 end root
−24–√3
negative 2 root index 3 start root 4 end root
−24−−−−√3 can be rewritten as -2√3
The answer is: −23–√3
To rewrite the irrational cube root −24−−−−√3 as a product of an integer and another irrational cube root, we need to simplify it. Here's how you can do it:
First, let's break down the given irrational cube root: −24−−−−√3.
To simplify the cube root, start by identifying perfect cube factors. In this case, we can see that 8 (2^3) is a perfect cube factor of 24. Therefore, we can rewrite the cube root as:
= -2√(3∙8)
Next, simplify the expression under the square root sign:
= -2√(3∙2^3)
Now, separate the perfect cube factor (2^3) from the remaining factor (√3):
= -2∙2∙√3
= -4√3
Therefore, the irrational cube root −24−−−−√3 can be rewritten as -4√3, which is the product of -4 (an integer) and √3 (another irrational cube root).