Answer:
a=k1
![\sqrt[3]{B}](https://img.qammunity.org/2021/formulas/mathematics/high-school/j6nykctg8hqb79x1nszfkq35mpyj3pzim9.png)
B=125
Explanation:
Given :
a=3
B=64
According to question
a ∝
![\sqrt[3]{B}](https://img.qammunity.org/2021/formulas/mathematics/high-school/j6nykctg8hqb79x1nszfkq35mpyj3pzim9.png)
therefore
a=k1
........Eq(1)
K1=
......Eq(2)
Putting the value of a and B we get in Eq(2) we get
![K1=(3)/(4)](https://img.qammunity.org/2021/formulas/mathematics/high-school/elacxso4j7gl2fgxvj985w5rsmgtyp5zn4.png)
Putting the value of k1 in Eq(1)
.......................Eq(3)
putting the value of a=15/4 IN Eq(3) we get
![(15)/(4)\ =(3)/(4) \sqrt[3]{B} \\\\\sqrt[3]{B}\ =\ 5\\Cubing\ both\ side\ we\ get\\B=125](https://img.qammunity.org/2021/formulas/mathematics/high-school/yzg2nh9yqc65nx1oca0mfv65tafa4aeh1c.png)