What is the square root 6n carried to the power of 2/3 in radical form
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If we consider "a" as the edge length, and "D" the cube's diagonal, we have that the square cube's diagonal is equal to the edge length's square plus the side diagonal (d) square (Pythagoras theorem)
a² + d² = D²
And since:
d² = a² + a²
Clearing a, we have:
a² = D²-d²
<span>a² = D²-2a²
</span><span>3a² = D²
</span>a = √(<span>D²/3)
</span>
Surface area is equal to 6·a², so the surface area will be 6·(D²/3) = 2D²
The volume is a³, so the volume will be √(D²/3)³ = √(
<span>/3</span>³<span>) = D</span>³/√27
a² + d² = D²
And since:
d² = a² + a²
Clearing a, we have:
a² = D²-d²
<span>a² = D²-2a²
</span><span>3a² = D²
</span>a = √(<span>D²/3)
</span>
Surface area is equal to 6·a², so the surface area will be 6·(D²/3) = 2D²
The volume is a³, so the volume will be √(D²/3)³ = √(