Question

Find the value of
A.\[\sqrt[3]{{392}} \times \sqrt[3]{{448}}\]
B.\[\sqrt[3]{{3375}} \times 729\]
C.Find the smallest number by which 17496 can be divided so that the quotient is a perfect cube. Also find the cube root of the quotient.

Answer 1

Hint: In the first question we will take the respective cube roots. Then the product of the roots will be the answer. In the second part we will use prime factorization to find the numbers that are perfect cubes and then the missing or left number will be our smallest number. As we do this in finding the square root, the same will be used for the cube root.

Complete step-by-step answer:
A.
\[\sqrt[3]{{392}} \times \sqrt[3]{{448}}\]
We know that,
i. \[\sqrt[3]{{392}} = {\left( {392} \right)^{\dfrac{1}{3}}}\]
Now we will write the number inside the bracket in prime factors form.
\[ = {\left( {2 \times 2 \times 2 \times 7 \times 7} \right)^{\dfrac{1}{3}}}\]
We can write 2 as cube form and 7 in square form with separate brackets,
\[ = {\left( {{2^3}} \right)^{\dfrac{1}{3}}} \times {\left( {{7^2}} \right)^{\dfrac{1}{3}}}\]
Now we know that \[{\left( {{a^m}} \right)^n} = {a^{m \times n}}\] using this property we can write,
\[ = {2^{3 \times \dfrac{1}{3}}} \times {7^{2 \times \dfrac{1}{3}}}\]
On calculating we get,
\[ = 2 \times {7^{\dfrac{2}{3}}}\]
Thus,
\[\sqrt[3]{{392}} = 2 \times {7^{\dfrac{2}{3}}}\]

We know that
ii. \[\sqrt[3]{{448}} = {\left( {448} \right)^{\dfrac{1}{3}}}\]
Now we will write the number inside the bracket in prime factors form.
\[\sqrt[3]{{448}} = {\left( {2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7} \right)^{\dfrac{1}{3}}}\]
We can write 2 as power form and 7 in single number form with separate brackets,
\[ = {\left( {{2^6}} \right)^{\dfrac{1}{3}}} \times {7^{\dfrac{1}{3}}}\]
Now we know that \[{\left( {{a^m}} \right)^n} = {a^{m \times n}}\] using this property we can write,
\[ = {2^{6 \times \dfrac{1}{3}}} \times {7^{\dfrac{1}{3}}}\]
On calculating we get,
\[ = {2^2} \times {7^{\dfrac{1}{3}}}\]
Thus,
\[\sqrt[3]{{448}} = {2^2} \times {7^{\dfrac{1}{3}}}\]
now from the given question we can write,
\[\sqrt[3]{{392}} \times \sqrt[3]{{448}} = 2 \times {7^{\dfrac{2}{3}}} \times {2^2} \times {7^{\dfrac{1}{3}}}\]
Taking the same bases nearer,
\[\sqrt[3]{{392}} \times \sqrt[3]{{448}} = 2 \times {2^2} \times {7^{\dfrac{2}{3}}} \times {7^{\dfrac{1}{3}}}\]
and using the law \[{a^m} \times {a^n} = {a^{m + n}}\],
Now adding the powers,
\[ = {2^3} \times {7^1}\]
We know the cube of 2 is 8 and any number to the power 1 is the number itself.
\[ = 8 \times 7\]
Thus on taking the product we get,
\[\sqrt[3]{{392}} \times \sqrt[3]{{448}} = 56\]

B.\[\sqrt[3]{{3375}} \times 729\]
Here we can use the same approach used above but since we know that 3375 is the perfect cube of 15 we will directly take the answer,
\[\sqrt[3]{{3375}} \times 729 = 15 \times 729\]
Taking the product we get,
\[\sqrt[3]{{3375}} \times 729 = 10935\]

C.Now we will solve the second question. For that we will express the given number in prime factors form.
\[17496 = 2 \times 8748\]
Now this will continue until we get all the number in product as prime numbers,
\[17496 = 2 \times 2 \times 4374\]
Again we will use 2,
\[17496 = 2 \times 2 \times 2 \times 2187\]
Now we will use 3 for splitting further,
\[17496 = 2 \times 2 \times 2 \times 3 \times 729\]
Now we either can use the same procedure or can say that 729 is the perfect cube of 9.
\[17496 = 2 \times 2 \times 2 \times 3 \times {9^3}\]
So now we can write as,
\[17496 = {2^3} \times 3 \times {9^3}\]
Now 3 is the number that does not have a cube form. So this is the smallest number that can be used to divide.
So the smallest number is 3.
now the cube root of quotient will be,
\[2 \times 9 = 18\]

Note: Note that we can continue the prime factorization for the second problem but we will reach the same answer. The smallest number is 3 because that is the only single number remaining and the others are in cube form. If asked for square form we will pair the numbers in terms of square.
talking about the first part we generally get confused in writing the radicals or roots. when the form above is given that can be written in indices form with fraction having 1 in numerator and that number in the denominator.
also note that either numbers with the same base can be combined or with the same power. not a different base or power number can be comb.