Question

If x= a +b+c;y=b+c+d where a, b,
c, d are the first four prime numbers then
(x + y)^2 (y^2 -x^2) is​?

Answer 1

Answer:

as per your problem

the first four prime numbers are 2 , 3 , 5 , 7

let a = 2

b = 3

c = 5

d = 7

then X = a+b+c = 10

Y. = b+c+d = 15

by the problem ( X + Y ) ^2 ( Y^2 - x^ 2) =

625 × ( 225 - 100 )

625 × 125 = 78125

Answer 2

(x + y)² (y² - x²) = 78125

Step-by-step explanation:

Since a, b, c, d are the first four Prime Numbers, then

• a = 2
• b = 3
• c = 5
• d = 7

Now, x = a + b + c = 2 + 3 + 5 = 10

and y = b + c + d = 3 + 5 + 7 = 15

So, x² = (10)² = 100

and y² = (15)² = 225

Then, y² - x² = 225 - 100 = 125

Also, x + y = 10 + 15 = 25

So, (x + y)² = (25)² = 625

Therefore, (x + y)² (y² - x²)

= 625 * 125

= 78125

#SPJ3