- #1

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_{T}and II

_{T}with those in 1.269. Could someone please explain? Thanks

with 1.269 being:

- I
- Thread starter FluidStu
- Start date

- #1

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with 1.269 being:

- #2

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$$2II_TA-2I_TB-A^3+3AB=2II_TA-2AB-A^3+3AB=A\left(2II_T-2B-A^2+3B\right)=A\left(2II_T+B-A^2\right)=2A\left(II_T-0.5(A^2-B)\right)

=2A\left(II_T-II_T\right)=0$$

So twice the difference is zero.

So the two RHSs are equal.

- #3

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Great! Thanks Andrew.

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