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The distance covered by a train in $(5 y-1)$ hours is $\left(125 y^3-1\right)$ km. The speed of the train is: |
$\left(5 y^3-1\right)$ km/h $\left(25 y^2-5 y+1\right)$ km/h $(5y+1)$ km/h $\left(25 y^2+5 y+1\right)$ km/h |
$\left(25 y^2+5 y+1\right)$ km/h |
Distance = Speed × Time Speed =\(\frac{125y^3 - 1 }{(5y - 1)}\) By using formula :- a^3 - b^3 = (a-b) ( a^2 + b^2 + ab) Speed = ( 25y^2 + 5y + 1 ) km/h |
