Question

Find the cube fo the following numbers by column method:
(i) 35 (ii) 56 (iii) 72

Answer 1

(i)$(35{)}^3\begin{array}{cccc}\text{I column}& \text{II column}& \text{III column}& \text{IV column}\\ a^3& 3\times a^2\times b& 3\times a\times b^2& b^3\\ (3{)}^3& 3\times (3{)}^2\times 5& 3\times 3\times (5{)}^2& (5{)}^3\\ =27& {}^{=}135& =225& =125\\ \frac{+15}{42}& \frac{+23}{158}& \frac{+12}{237}& \\ 42& 8& 7& 5\end{array}\therefore (35{)}^3=42875$
(ii) $(56{)}^3\begin{array}{cccc}\text{I column}& \text{II column}& \text{III column}& \text{IV column}\\ a^3& 3\times a^2\times b& 3\times a\times b^2& b^3\\ (5{)}^3& 3\times 5^2\times 6& 3\times 3\times 6^2& (6{)}^3\\ =125& =450& =540& =216\\ \frac{+50}{175}& \frac{+56}{506}& \frac{+21}{561}& \\ 175& 6& 1& 6\end{array}\therefore (56{)}^3=175616$
(iii) $(72{)}^3\begin{array}{cccc}\text{I column}& \text{II column}& \text{III column}& \text{IV column}\\ a^3& 3\times a^2\times b& 3\times a\times b^2& b^3\\ (7{)}^3& 3\times 7^2\times 2& 3\times 7\times 2^2& (2{)}^3\\ =343& =294& =84& =8\\ \frac{+30}{373}& \frac{+8}{302}& & \\ 373& 2& 4& 8\end{array}\therefore (72{)}^3=373248$