document.write( "Question 35101: using deductive methods prove that if n is an interger, n is greater than or equal to 2 , then f(n)= n cubed - n is always divisible by at least 6 \n" ); document.write( "

Algebra.Com's Answer #21409 by venugopalramana(3286) $\"\"$ $\"About$

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F(N)=N^3-N=N(N^2-1)=N(N+1)(N-1)=(N-1)N(N+1)=PRODUCT OF 3 CONSECUTIVE INTEGERS.
\n" ); document.write( "SO ONE OF THEM IS DEFINITELY DIVISIBLE BY 2.
\n" ); document.write( "FURTHER WHEN WE DIVIDE A NUMBER WITH 3 WE CAN GET A REMAINDER OF 1 OR 2 OR 0 THAT IS DIVISIBLE BY 3.SO SINCE THIS IS A PRODUCT OF 3 CONSECUTIVE INTEGERS,ONE OF THEM HAS TO BE DIVISIBLE BY 3.
\n" ); document.write( "SO WE GET THAT THE PRODUCT IS DIVISIBLE BY 2 AS WELL AS 3.HENCE THE PRODUCT IS DIVISIBLE BY 6 \n" ); document.write( "

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