Question

if a cos³α + 3a cos α sin²α=m and a sin³α + 3a cos²α sin α=n. Then (m+n)⅔ + (m-n)⅔ is equal to :-

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Answer 1

Answer:

Step-by-step explanation:

the first step is to add m+n then

a*cos^3(α)+3a cos(α)sin^2(α)=m

asin^3(α)+3acos^2(α)sin(α)=n

then (m+n)=a(sin(α)+cos(α))^3

hence similarly

(m-n)=a(sin(α)-cos(α))^3

hence [(m+n)^(2/3)+(m-n)^(2/3)=2*a^(2/3)]

Answer 2

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$\red{2 {a}^{ \frac{2}{3} }}$

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Step 1 => Adding m amd n (m + n)

a cos³α + 3a cos α sin²α + a sin³α + 3a cos²α sin α = m + n

taking common a

a [ cos³α+ 3 cos α sin²α + sin³α + 3 cos²α sin α] = m + n

a[cos α + sin α]³ = m + n — — — — (i)

Step 2 => On subtracting n from m (m - n) we get

a[cos α - sin α]³ = m - n — — — — (ii)

Step 3 => Substituting the value of eq. (i) and eq. (ii)

a⅔ [cos α + sin α]² + a⅔ [cos α - sin α]²

taking common a⅔

a⅔ { [cos α + sin α]² + [cos α - sin α]² }

a⅔ { 1 + 2sin α cos α + 1 - 2sin α cos α }

$\large {\boxed{ \bold{2 {a}^{ \frac{2}{3} } }}}$

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