Question

In any △ ABC prove that ,
$\sf \: a³ \: sin (B - C) + b³ \: sin (C - A) + c³ \: sin (A - B) = 0$

Answer 1

LHS

=a³sin(B−C)

=(2RsinA)³sin(B−C)

=2R³.2sin²A⋅2sinA⋅sin(B−C)

=2R³⋅(1−cos2A)⋅[cos(A−B+C)−cos(A+B−C)]

=2R³⋅(1−cos2A)⋅[cos(π−2B)−cos(π−2C)]

=2R³⋅(1−cos2A)⋅(cos2C−cos2B)

=2R³⋅(cos2C−cos2B−cos2A⋅cos2C+cos2A⋅cos2B)

Similarly, the second part

=2R³ (cos2A−cos2C−cos2A⋅cos2B+cos2B⋅cos2C)

And the third part

=2R³ (cos2B−cos2A−cos2B⋅cos2C+cos2A⋅cos2C)

Adding up all these three parts, we get,

a³sin(B−C)+b³sin(C−A)+c³sin(A−B)=0

Hence proved

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Answer 2

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm \: {a}^{3}sin(B - C) + {b}^{3}sin(C - A) + {c}^{3}sin(A - B)$

can be rewritten as

$\rm \: = {a}^{2}(asin(B - C)) + {b}^{2}(bsin(C - A)) + {c}^{2}(csin(A - B))$

We know, from Sine Law

In △ ABC,

$\boxed{\tt{ \rm \: \: \frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC} = k \: }}$

So, from this, we get

$\rm \: a = k \: sinA$

$\rm \: b = k \: sinB$

$\rm \: c = k \: sinC$

So, on substituting the values, we get

$\rm \: = {a}^{2}ksinAsin(B - C) + {b}^{2}ksinBsin(C - A) + {c}^{2}ksinCsin(A - B)$

can be further rewritten as

$\rm \: =k({a}^{2}sinAsin(B - C) + {b}^{2}sinBsin(C - A) + {c}^{2}sinCsin(A - B))$

Let we Consider

$\rm \: sinA = sin[\pi - (B + C)] = sin(B + C)$

So,

$\rm \: sinAsin(B - C) = sin(B + C)sin(B - C) = {sin}^{2}B - {sin}^{2}C$

Thus,

$\rm \: sinAsin(B - C) = {sin}^{2}B - {sin}^{2}C$

Similarly,

$\rm \: sinBsin(C - A) = {sin}^{2}C - {sin}^{2}A$

and

$\rm \: sinCsin(A - B) = {sin}^{2}A - {sin}^{2}B$

So, on substituting these values, we get

$\rm \: = k[ {a}^{2}({sin}^{2}B - {sin}^{2}C) + {b}^{2}({sin}^{2}C - {sin}^{2}A) + {c}^{2}({sin}^{2}A - {sin}^{2}B)$

On substituting the values of a, b and c, from sine law, we get

$\rm \: = k[ {k}^{2} {sin}^{2}A ({sin}^{2}B - {sin}^{2}C) + {k}^{2} {sin}^{2}B({sin}^{2}C - {sin}^{2}A) + {k}^{2} {sin}^{2}C({sin}^{2}A - {sin}^{2}B)]$

$\rm \: = {k}^{3} [{sin}^{2}A ({sin}^{2}B - {sin}^{2}C) +{sin}^{2}B({sin}^{2}C - {sin}^{2}A) +{sin}^{2}C({sin}^{2}A - {sin}^{2}B)]$

$\rm \: = {k}^{3}( {sin}^{2}A {sin}^{2}B - {sin}^{2}A {sin}^{2}C + {sin}^{2}B {sin}^{2}C - {sin}^{2}B {sin}^{2}A + {sin}^{2}C {sin}^{2}A - {sin}^{2}C {sin}^{2}B)$

$\rm \: = \: {k}^{3} \times 0$

$\rm \: = \: 0$

Hence,

$\boxed{\tt{ \rm \: {a}^{3}sin(B - C) + {b}^{3}sin(C - A) + {c}^{3}sin(A - B) = 0 \: }}$

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ADDITIONAL INFORMATION

1. Cosine Law

$\boxed{\tt{ \: \: cosA = \frac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc} \: \: }}$

$\boxed{\tt{ \: \: cosB = \frac{ {c}^{2} + {a}^{2} - {b}^{2} }{2ca} \: \: }}$

$\boxed{\tt{ \: \: cosC = \frac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab} \: \: }}$

2. Projection Formula

$\boxed{\tt{ a = bcosC + ccosB \: }}$

$\boxed{\tt{ b = acosC + ccosA \: }}$

$\boxed{\tt{ c = acosB + bcosA \: }}$