Question

If x and y are two positive real numbers such that 9x²+4y²=97 then find the value of 27x³+64y³

Answer 1

we can find the value of x&y by hit and trial method

sum of two squares is 97

81+16=97

there for x^2=9
x=3

y^2= 4
y=2

27x^3+64y^3
27(3)^3+64(2)^3
27x27+64x8
729+ 512
1241 answer

Answer 2

Answer:

27x³+64y³  = 1241

Step-by-step explanation:

Given problem

If x and y are two positive real numbers such that 9x²+4y²=97 then find the value of 27x³+64y³    

⇒    9x²+4y² = 97  

⇒    (3x)² + (2y)² = 97 _(1)

given that x and y are two positive real numbers  

⇒ now find out for which values (1) will satisfy  

[ we can find these values with trial and error method ]

⇒ at x = 3 and y = 2   (1) will satisfy  

⇒   3² (3²) + 2² (2)² = 9(9) + 4(4) = 81 + 16 = 97

now substitute x = 3 and y = 2 values in 27x³+64y³    

27x³+64y³  =  27(3)³ + 64(2)³

                  = 27 (27) + 64 (8)

                  = 729 + 512

                  = 1241