Question

if a+b+c=0, then show that a³+b³+c³=3abc​

Answer 1

Answer:

Before starting with the solution of this question, let us understand the concept.

\begin{gathered}\\\end{gathered}

Concept used:

cos also known as 'cosine' is the complimentary of sine as the name suggests.

So,

cos θ = sin (90 - θ)

\begin{gathered}\\\end{gathered}

Step-by-step explanation:

Applying this concept,

cos 72° can be written as sin (90° - 72°)

→ cos 72° = sin (90° - 72°)

⇒ cos 72° = sin 18°

\begin{gathered}\\\end{gathered}

Now, dividing by cos 72° on both sides,

\begin{gathered} \sf{ \dfrac{cos \: 72^{ \circ} }{cos \: 72 ^{ \circ} } = \dfrac{sin \: {18}^{ \circ} }{ cos \: 72 ^{ \circ} } } \\ \\ \end{gathered}

cos72

∘

cos72

∘

=

cos72

∘

sin18

∘

\begin{gathered} \implies \: \sf{ \dfrac{ \cancel{cos \: 72^{ \circ}} } { \cancel{cos \: 72 ^{ \circ}} } = \dfrac{sin \: {18}^{ \circ} }{ cos \: 72 ^{ \circ} } } \\ \\ \end{gathered}

⟹

cos72

∘

cos72

∘

=

cos72

∘

sin18

∘

\begin{gathered} \implies \sf{ 1= \dfrac{sin \: {18}^{ \circ} }{ cos \: 72 ^{ \circ} } } \\ \\ \end{gathered}

⟹1=

cos72

∘

sin18

∘

\begin{gathered} \therefore \: \boxed{ \bf{\dfrac{sin \: {18}^{ \circ} }{ cos \: 72 ^{ \circ} }} = 1} \\ \\ \end{gathered}

∴

cos72

∘

sin18

∘

=1

Answer 2

Answer:

by identity Viii

We know that,

x3+y3+z3-3xyz = (x +y+z)(x2+y2+z2–xy–yz–xz)

Now, according to the question, let (x+y+z) = 0,

then, x3+y3+z3 -3xyz = (0)(x2+y2+z2–xy–yz–xz)

⇒ x3+y3+z3–3xyz = 0

⇒ x3+y3+z3 = 3xyz

Hence Proved

Step-by-step explanation:

hope this will help you