Question

The difference in volumes of two cubes is 152 m³ and the difference in their one face areas is 20 m². If the sum of their edges is 10 m, the product of the edges is:

Answer 1

Answer:24 m

Explanation

Given

Difference of two cubes=152 m^3

difference in one face area=20 m^2

Sum of their edges=10 m

let the edges of the two cubes be x and y m

Then,

x^3-y^3=152

and x^2-y^2=20

Also x+y=10

so

x-y=(x^2-y^2)/(x+y)

=20/10

=2

Now,

(x^3-y^3)/(x-y)

152/2

76

x^2+y^2+xy=76

(x+y)^2-xy=76

xy=(x+y)^2-76

xy=(10)^2-76

xy=100-76

xy=24

______________________

Therefore the product of the edges is 24

Answer 2

Answer:

The product of the edges is 24m

Step-by-step explanation:

Given Problem:

The difference in volumes of two cubes is 152 m³ and the difference in their one face areas is 20 m². If the sum of their edges is 10 m, the product of the edges is?

Solution:

Given that,

Difference of two cubes = 152 m³

Difference in one face area =20 m²

Sum of their edges = 10 m

Let the edges of the two cubes be 'a' and 'b' meter,

So,

$\sf => a^3-b^3=152\\\\\ =>a^2-b^2=20\\\\\ =>a+b=10\\\\\bf {So,}\\\\\sf => a-b=\dfrac{(a^2-b^2)}{(a+b)}\\\\\ =>\dfrac{20}{10}\\\\\ =>2\\\\ \bf Now,\\\\\sf =>\dfrac{(a^3-b^3)}{(a-b)}\\\\\ =>\dfrac{152}{2}\\\\\ => 76\\\\\ =>a^2+b^2+ab=76\\\\\ =>(a+b)^2-ab=76\\\\\ =>ab=(a+b)^2-76\\\\\ =>ab=(10)^2-76\\\\\ =>ab=100-76\\\\\ =>ab=24\\\\\ Hence, the\; product\; of\; the\; edges\; is\; 24.$