Given that π = 3β 32π4βπ/2 . Show that π = 1/2 4β2π2 + π
T=32c4βq23=T3=32c4βq2=T3+q2=32c4=132(T3+q2)=c4=116β 12(2T3+q)=c4β ββΉβ βc=1212(2T3+q)4T = \sqrt[3]{32c^4-\frac{q}{2}}\\ =T^3 = 32c^4-\frac{q}{2}\\ =T^3+\frac{q}{2}=32c^4\\ =\frac{1}{32}(T^3+\frac{q}{2})=c^4\\ =\frac{1}{16}\cdot\frac{1}{2}(2T^3+q)=c^4\\ \implies c = \frac{1}{2}\sqrt[4]{\frac{1}{2}(2T^3+q)}T=332c4β2qββ=T3=32c4β2qβ=T3+2qβ=32c4=321β(T3+2qβ)=c4=161ββ 21β(2T3+q)=c4βΉc=21β421β(2T3+q)β
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