Given that π = 3β 32π4βπ /2 . Show that π = 1/2 4β2π3 + π
π=32π4βπ32π = \frac{\sqrt[3]{32π^4βπ }}{2}T=2332c4βqββ
2ΓT=32π4βq32ΓT=\sqrt[3]{32π^4-q}2ΓT=332c4βqβ
23ΓT3=32π4βq2^3ΓT^3={32π^4-q}23ΓT3=32c4βq
8T3=32π4βq8T^3={32π^4-q}8T3=32c4βq
8T3+q=32π48T^3+q={32π^4}8T3+q=32c4
\sqrt[4]\frac{8T^3+q}{32}={π}
c=122T3+q4c=\frac{1}{2}\sqrt[4]{2T^3+q}c=21β42T3+qβ
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