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Square Root and Cube-Cube Root — MATHS STD 8 — Question

Gujarat BoardEnglish MediumSTD 8MATHSSquare Root and Cube-Cube Root1 Mark
Question
For every natural number m, $(2m - 1, 2m^2- 2m, 2m^2- 2m + 1)$ is a pythagorean triplet.
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Answer

False.
Solution:
$ \because(2 m-1)^2 \neq\left(2 m^3-23\right)^2+\left(2 m^2-2 m+1\right)^2 $
$ \left(2 m^3-2 m\right)^2 \neq(2 m-1)^2+\left(2 m^2-2 m+1\right)^2 $
and $\left(2 m^2-2 m+1\right)^2 \neq(2 m-1)^2+\left(2 m^3-2 m\right)^2 $

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