Sure, let's address each question step-by-step.
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### Question 26:
Find the rationalizing factor of [tex]\(\sqrt[3]{9}\)[/tex].
To rationalize [tex]\(\sqrt[3]{9}\)[/tex], we need a factor which, when multiplied by [tex]\(\sqrt[3]{9}\)[/tex], results in a rational number.
The cube root of 9 is [tex]\(\sqrt[3]{9}\)[/tex]. To rationalize a cube root, we look for another cube root that, when multiplied by [tex]\(\sqrt[3]{9}\)[/tex], yields a perfect cube (i.e., a rational number).
Consider [tex]\(\sqrt[3]{9} \times \sqrt[3]{3} = \sqrt[3]{27} = 3\)[/tex].
So, the rationalizing factor here is [tex]\(\sqrt[3]{3}\)[/tex].
Thus, the answer is:
a) [tex]\(\sqrt[3]{3}\)[/tex]
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### Question 27:
Simplify [tex]\(\frac{a^8 - b^8}{(a^4 + b^4)(a^2 + b^2)}\)[/tex] given [tex]\(a \neq 0\)[/tex] and [tex]\(b \neq 0\)[/tex].
We start with the numerator:
[tex]\[ a^8 - b^8 \][/tex]
This expression can be factored using the difference of powers formula:
[tex]\[ a^8 - b^8 = (a^4 - b^4)(a^4 + b^4) \][/tex]
Next, factor [tex]\(a^4 - b^4\)[/tex]:
[tex]\[ a^4 - b^4 = (a^2 - b^2)(a^2 + b^2) \][/tex]
Thus,
[tex]\[ a^8 - b^8 = (a^2 - b^2)(a^2 + b^2)(a^4 + b^4) \][/tex]
Now, substituting this into our original fraction:
[tex]\[ \frac{a^8 - b^8}{(a^4 + b^4)(a^2 + b^2)} = \frac{(a^2 - b^2)(a^2 + b^2)(a^4 + b^4)}{(a^4 + b^4)(a^2 + b^2)} \][/tex]
Notice that [tex]\( (a^4 + b^4) \)[/tex] and [tex]\( (a^2 + b^2) \)[/tex] cancel out:
[tex]\[ \frac{(a^2 - b^2)\cancel{(a^2 + b^2)}\cancel{(a^4 + b^4)}}{\cancel{(a^4 + b^4)}\cancel{(a^2 + b^2)}} = a^2 - b^2 \][/tex]
Thus, the simplified expression is:
c) [tex]\(a^2 - b^2\)[/tex]
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So, the answers are:
- For Question 26: a) [tex]\(\sqrt[3]{3}\)[/tex]
- For Question 27: c) [tex]\(a^2 - b^2\)[/tex]
