`(25^2-15^2)^((3)/(2))=` _____

(25.732)^(2)-(15.732)^(2)=?

((15.4)^(2)-(35.4)^(2))/(25)=

Sum to n terms of the series 1^(3) - (1.5)^(3) +2^(3)-(2.5)^(3) +…. is

Sum to n terms of the series 1^(3) - (1.5)^(3) +2^(3)-(2.5)^(3) +…. is

If a^(2)+9b^(2)+25c^(2)=abc((15)/(a)+(5)/(b)+(3)/(c)), then a,b,c are in

(3,4,5),(5,12,13),(8,15,17) etc.are Pythagorea triplets,because 3^(2)+4^(2)=25=5^(2)5^(2)+12^(2)=169=13^(2)8^(2)+15^(2)=289=17^(2)

(10x^(2)+15x+63)/(5x^(2)-25x+12)=(2x+3)/(x-5)

Solve the following equation: (10x^2+15x+63)/(5x^2-25x+12)=(2x+3)/(x-5)

Prove that: (3^(-3)x6^(2)x sqrt(98))/(5^(2)x(1)/(25)3x(15)^(-(4)/(3))x3^((1)/(3)))=28sqrt(2)