Question

Using suitable identity find 41² - 39³.​

Answer 1

Answer:

Using (a+b)

2

= a

2

+2ab+b

2

Writing (41)

2

as (40+1)

2

= 40

2

+2×40+1+1

2

= 1600+80+1

The solution is 1681

Answer 2

Answer:

$(40+1)^{2}$

$(30+9)^{2} \\(a + b)^{2} = a^{2} + 2ab + b^{2} \\here a=30,b=9\\(30+9)^{2} =30^{2} +2ab+9^{2} \\=900+2*30*9+81\\=900+540+81\\=1521$

$USING IDENTITY:-\\(a + b)^{2} = a^{2} + 2ab + b^{2} \\ Here /a = 40, b = 1(40 + 1)^{2} = 40^{2} + 2 + (1)^{2} = 1600 + 80 + 1$

$1681+-1521\\=160$

Step-by-step explanation:

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