Given that n n\,n is an integer, show that
(3n−1)3=9(3n3−3n2+n)−1 (3n-1)^3=9(3n^3-3n^2+n)-1 (3n−1)3=9(3n3−3n2+n)−1and
(3n+1)3=9(3n3+3n2+n)+1 (3n+1)^3=9(3n^3+3n^2+n)+1 (3n+1)3=9(3n3+3n2+n)+1Hence prove that all cube numbers are either a multiple of 9 or 1 more or one less than a multiple of 9.
257 exam-style questions on Edexcel A Level Maths Algebraic Methods, covering 7.1 Algebraic Fractions, 7.2 Dividing Polynomials, 7.3 The Factor Theorem, 7.4 Mathematical Proof, and 7.5 Methods of Proof. Each one has a worked solution and a mark scheme showing where the marks go.