What is the irrational cube root 40−−√3 rewritten as a product of an integer and another irrational cube root?(1 point)
Responses
85–√3
8 root index 3 start root 5 end root
210−−√3
2 root index 3 start root 10 end root
−25–√3
negative 2 root index 3 start root 5 end root
25–√3
none of the above
What is the irrational cube root 40−−√3 rewritten as a product of an integer and another irrational cube root?(1 point)
Responses
85–√3
8 root index 3 start root 5 end root
210−−√3
2 root index 3 start root 10 end root
−25–√3
negative 2 root index 3 start root 5 end root
25–√3
8 root index 3 start root 5 end root
To rewrite the irrational cube root (√3) of 40 as a product of an integer and another irrational cube root, you can follow these steps:
Step 1: Simplify the cube root of 40:
The cube root of 40 can be simplified as the cube root of 8 times the cube root of 5, since 40 can be expressed as 8 multiplied by 5:
∛40 = ∛(8 × 5) = ∛8 × ∛5
Step 2: Express the cube root of 8 as an integer times the cube root of another number:
The cube root of 8 can be written as 2 times the cube root of 4, since 8 is equal to 2 multiplied by 4:
∛8 = 2 × ∛4
Step 3: Combine the expressions from step 2 and step 1:
∛40 = (∛8 × ∛5) = (2 × ∛4) × (∛5)
Therefore, the irrational cube root of 40 rewritten as a product of an integer and another irrational cube root is:
2∛4 × ∛5