Question

can any solve this pls​

Answer 1

Answer:

Step-by-step explanation:

acos³Ф+3acosФsin²Ф = m  asin³Ф+3asinФcos²Ф = n

(m+n)to the power2/3+(m-n) to the power 2/3

= (acos³Ф+3acosФsin²Ф+asin³Ф+3acos²ФsinФ)to the power2/3 +       (acos³Ф+3acosФsin²Ф-asin³Ф+3acos²ФsinФ)to the power2/3

formula  a³+b³+3ab(a+b)  a has taken comman

= a to the power2/3 ( cos³Ф+sin³Ф+3sinФcosФ(sinФ+cosФ) +                        a to the power2/3 ( cos³Ф-sin³Ф+3sinФcosФ(sinФ-cosФ)

again a has taken common

= a to the power2/3(((cosФ+sinФ)³) to the power2/3 + ((cosФ-sinФ)³)          to the power2/3)          ( 3 cancelled)

= a to the power2/3 ((cosФ+sinФ)²+(cosФ-sinФ)²)

=a to the power2/3(sin²Ф+cos²Ф+2sinФcosФ+sin²Ф+cos²Ф-2sinФcosФ)

= a to the power2/3( 1+1)

= 2a to the power2/3