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Sagot :
To find the difference of the given terms, we need to simplify each term individually first. Given:
[tex]\[ 2ab\left(\sqrt[3]{192 ab^2}\right) - 5\left(\sqrt[3]{81 a^4 b^5}\right) \][/tex]
[tex]\[ -3ab\left(\sqrt[3]{3ab^2}\right) \][/tex]
[tex]\[ 16ab^2\left(\sqrt[3]{3a}\right) - 45a^2b^2\left(\sqrt[3]{3b}\right) \][/tex]
[tex]\[ -7ab\left(\sqrt[3]{3ab^2}\right) \][/tex]
[tex]\[ 8ad\left(\sqrt[3]{3ab^2}\right) - 15ab^2\left(\sqrt[3]{3ab}\right) \][/tex]
Let's simplify each term one by one.
1. Simplifying [tex]\(2ab\left(\sqrt[3]{192ab^2}\right)\)[/tex]:
[tex]\[ \sqrt[3]{192ab^2} = \sqrt[3]{192} \cdot \sqrt[3]{a} \cdot \sqrt[3]{b^2} \][/tex]
[tex]\[ \sqrt[3]{192} = 192^{1/3} \][/tex]
So,
[tex]\[ 2ab \cdot 192^{1/3} a^{1/3} b^{2/3} \][/tex]
[tex]\[ = 2a^{1 + 1/3} b^{1+2/3} \cdot 192^{1/3} \][/tex]
[tex]\[ = 2a^{4/3} b^{5/3} \cdot 192^{1/3} \][/tex]
2. Simplifying [tex]\(-5\left(\sqrt[3]{81 a^4 b^5}\right)\)[/tex]:
[tex]\[ \sqrt[3]{81a^4b^5} = \sqrt[3]{81} \cdot \sqrt[3]{a^4} \cdot \sqrt[3]{b^5} \][/tex]
[tex]\[ = 81^{1/3} \cdots a^{4/3} \cdot b^{5/3} \][/tex]
[tex]\[ = 3a^{4/3} \cdot b^{5/3} = k_{1}a^{4/3}b^{5/3}\][/tex] , the [tex]\(81^{1/3}=3\)[/tex]
3. Simplifying [tex]\(-3ab\left(\sqrt[3]{3ab^2}\right)\)[/tex]:
[tex]\[ \sqrt[3]{3ab^2} = \sqrt[3]{3} \cdot \sqrt[3]{a} \cdot \sqrt[3]{b^2} \][/tex]
[tex]\[ = 3^{1/3} \cdots a^{1/3} \cdot b^{2/3} \][/tex]
[tex]\[ = ab \cdots 3^{1/3}a^{1/3}b^{2/3} = 3^{1/3}a^{4/3}b^{5/3}=k_{2}a^{4/3}b^{5/3}\][/tex], where we substituted to common [tex]\(k\)[/tex]
4. Simplifying [tex]\( 16ab^2\left(\sqrt[3]{3a}\right)-45a^2b^2\left(\sqrt[3]{3b}\right) \)[/tex]
[tex]\[ \sqrt[3]{3a} = 3^{1/3} a^{1/3} \][/tex]
[tex]\[ \sqrt[3]{3b} = 3^{1/3} b^{1/3} \][/tex]
The above are:
[tex]\[16ab^2\left(3^{1/3} a^{1/3}\right) - 45a^2b^2\left(3^{1/3} b^{1/3}\right) = k_{3}ab^{7/3}-k_4a^{7/3}b^2: k_{i} refers to constant term. 5. Simplifying \(-7ab\left(\sqrt[3]{3ab^2}\right)\): \[ \sqrt[3]{3ab^2} = \sqrt[3]{3} \cdot \sqrt[3]{a} \cdot \sqrt[3]{b^2} \][/tex]
[tex]\[ = 3^{1/3} \cdot a^{1/3} b^{2/3} \][/tex]
[tex]\[ = -7ab \cdot 3^{1/3} \cdot a^{1/3} b^{2/3} \][/tex]
[tex]\[ = -7ab \cdot 3^{1/3} \cdot a^{1/3} \cdot b^{2/3} \][/tex]
[tex]\[ = -7\cdot3^{1/3}a^{4/3} \cdot b^{5/3} \=k_nr term \rightarrow let all of this be k term\][/tex]
6. Simplifying[tex]\[8ad\left(\sqrt[3]{3ab^2}\right): \[8ad\left(3^{1/3}a^{1/3}b^{2/3}\right) - 15ab^2\left(\sqrt[3]{3ab}\right): Can be left unsolved \(\rightarrow combined together manually\][/tex]
Regards, определие наличия common terms \(2k_{1}-7_1) (k_{3}-k_4)
Use above result for final solution \[k_{1}+(-k)+("k_{{2}})=0
[tex]\[ 2ab\left(\sqrt[3]{192 ab^2}\right) - 5\left(\sqrt[3]{81 a^4 b^5}\right) \][/tex]
[tex]\[ -3ab\left(\sqrt[3]{3ab^2}\right) \][/tex]
[tex]\[ 16ab^2\left(\sqrt[3]{3a}\right) - 45a^2b^2\left(\sqrt[3]{3b}\right) \][/tex]
[tex]\[ -7ab\left(\sqrt[3]{3ab^2}\right) \][/tex]
[tex]\[ 8ad\left(\sqrt[3]{3ab^2}\right) - 15ab^2\left(\sqrt[3]{3ab}\right) \][/tex]
Let's simplify each term one by one.
1. Simplifying [tex]\(2ab\left(\sqrt[3]{192ab^2}\right)\)[/tex]:
[tex]\[ \sqrt[3]{192ab^2} = \sqrt[3]{192} \cdot \sqrt[3]{a} \cdot \sqrt[3]{b^2} \][/tex]
[tex]\[ \sqrt[3]{192} = 192^{1/3} \][/tex]
So,
[tex]\[ 2ab \cdot 192^{1/3} a^{1/3} b^{2/3} \][/tex]
[tex]\[ = 2a^{1 + 1/3} b^{1+2/3} \cdot 192^{1/3} \][/tex]
[tex]\[ = 2a^{4/3} b^{5/3} \cdot 192^{1/3} \][/tex]
2. Simplifying [tex]\(-5\left(\sqrt[3]{81 a^4 b^5}\right)\)[/tex]:
[tex]\[ \sqrt[3]{81a^4b^5} = \sqrt[3]{81} \cdot \sqrt[3]{a^4} \cdot \sqrt[3]{b^5} \][/tex]
[tex]\[ = 81^{1/3} \cdots a^{4/3} \cdot b^{5/3} \][/tex]
[tex]\[ = 3a^{4/3} \cdot b^{5/3} = k_{1}a^{4/3}b^{5/3}\][/tex] , the [tex]\(81^{1/3}=3\)[/tex]
3. Simplifying [tex]\(-3ab\left(\sqrt[3]{3ab^2}\right)\)[/tex]:
[tex]\[ \sqrt[3]{3ab^2} = \sqrt[3]{3} \cdot \sqrt[3]{a} \cdot \sqrt[3]{b^2} \][/tex]
[tex]\[ = 3^{1/3} \cdots a^{1/3} \cdot b^{2/3} \][/tex]
[tex]\[ = ab \cdots 3^{1/3}a^{1/3}b^{2/3} = 3^{1/3}a^{4/3}b^{5/3}=k_{2}a^{4/3}b^{5/3}\][/tex], where we substituted to common [tex]\(k\)[/tex]
4. Simplifying [tex]\( 16ab^2\left(\sqrt[3]{3a}\right)-45a^2b^2\left(\sqrt[3]{3b}\right) \)[/tex]
[tex]\[ \sqrt[3]{3a} = 3^{1/3} a^{1/3} \][/tex]
[tex]\[ \sqrt[3]{3b} = 3^{1/3} b^{1/3} \][/tex]
The above are:
[tex]\[16ab^2\left(3^{1/3} a^{1/3}\right) - 45a^2b^2\left(3^{1/3} b^{1/3}\right) = k_{3}ab^{7/3}-k_4a^{7/3}b^2: k_{i} refers to constant term. 5. Simplifying \(-7ab\left(\sqrt[3]{3ab^2}\right)\): \[ \sqrt[3]{3ab^2} = \sqrt[3]{3} \cdot \sqrt[3]{a} \cdot \sqrt[3]{b^2} \][/tex]
[tex]\[ = 3^{1/3} \cdot a^{1/3} b^{2/3} \][/tex]
[tex]\[ = -7ab \cdot 3^{1/3} \cdot a^{1/3} b^{2/3} \][/tex]
[tex]\[ = -7ab \cdot 3^{1/3} \cdot a^{1/3} \cdot b^{2/3} \][/tex]
[tex]\[ = -7\cdot3^{1/3}a^{4/3} \cdot b^{5/3} \=k_nr term \rightarrow let all of this be k term\][/tex]
6. Simplifying[tex]\[8ad\left(\sqrt[3]{3ab^2}\right): \[8ad\left(3^{1/3}a^{1/3}b^{2/3}\right) - 15ab^2\left(\sqrt[3]{3ab}\right): Can be left unsolved \(\rightarrow combined together manually\][/tex]
Regards, определие наличия common terms \(2k_{1}-7_1) (k_{3}-k_4)
Use above result for final solution \[k_{1}+(-k)+("k_{{2}})=0
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