Math, asked by Tharani6, 1 year ago

plz send with full method. today is my test

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Shanaya2504: which school r u in
Shanaya2504: i m also in 9th
Shanaya2504: and i m aslo having maths test today

Answers

Answered by mriganka2
3
1 )))))
 {a}^{3}  +  \frac{1}{ {a}^{3} }
 = (a +  \frac{1}{a} )^{3}  - 3 \times a \times  \frac{1}{a} (a +  \frac{1}{a} )
 =  {p}^{3}  - 3p \:  \: (as \:  \:  \:  \: a +  \frac{1}{a}  = p)
 = p( {p}^{2}  - 3) \:  \:  \:  \:  \:  \:  \:  \: proved.

2 )))))
 {a}^{3}  + 8 {b}^{3}  + 30ab
 =  {a}^{3}  + (2b)^{3}  + 3 \times a \times 2b  \times 5
 =  {a}^{3}  + (2b) ^{3}  + 3 \times a \times 2b \times (a + 2b)
 =  {(a + 2b)}^{3}
 =  {5}^{3}
 = 125 \:  \:  \:  \:  \:  \:  \: proved.


Tharani6: how u had done 2nd part 3rd step
Tharani6: plz tell
mriganka2: as we know (a+b)^3 = a^3+b^3+3ab(a+b)
Tharani6: yes so
Tharani6: so how u have done
mriganka2: so I input that formula in the sum
mriganka2: in the sum a=a and b=2b
Tharani6: how u have done a³+(2b)³+3×a×2b(a+b) = (a+2b)³
mriganka2: see the formula I have written there in the sum 'a' represents a but 'b' represents 2b
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