Math, asked by limbuamrita58, 2 days ago

find hcf of 30(a^2-9) and 75(a^3+27)​

Answers

Answered by yenkarprakash5
0

Step-by-step explanation:

18,48

18=2\times3\times318=2×3×3

48=2\times2\times2\times2\times348=2×2×2×2×3

The common factor of 18 and 48 are 2,3. thus, HCF of 18 and 48 is2\times3=62×3=6

ii) 30,42

30=2\times3\times530=2×3×5

42=2\times3\times742=2×3×7

The common factor of 30 and 42 are 2 and 3

thus, HCF of 30,42 are 2\times3=62×3=6

iii) 18,60

18=2\times3\times318=2×3×3

60=2\times2\times3\times560=2×2×3×5

The common factor of 18 and 60 are 2,3. thus, HCF of 18 and 60 are 2\times3=62×3=6

iv) 27, 63

27=3\times3\times327=3×3×363=3\times3\times763=3×3×7

The common factor for 27 and 63 is 3

thus, HCF of 27 and 63 is 3\times3=93×3=9

v) 36,84

36=2\times2\times3\times336=2×2×3×384=2\times2\times3\times784=2×2×3×7

The common factor of 36 and 84 is 2\times32×3

thus, HCF of 36 and 84 is 2\times2\times3=122×2×3=12

vi) 34, 102

34=2\times1734=2×17102=2\times3\times17102=2×3×17

The common factor in 34 and 102 is 2, 17.

thus, HCF for 34 and 102 is 2\times17=342×17=34

vi) 70, 105, 175

70=2\times5\times770=2×5×7

105=3\times5\times7105=3×5×7

175=5\times5\times7175=5×5×7

The common factor in 70, 105 and 175 are 5, 7.

thus, HCF for 70,105 and 175 is 5\times7=355×7=35

vii) 91,112,49

91=7\times1391=7×13112=2\times2\times2\times2\times7112=2×2×2×2×749=7\times749=7×7

The common factors of 91, 112, and 49 are 7.

therefore, HCF of 91, 112 and 49 are 7.

viii) 18, 54,81

18=2\times3\times318=2×3×354=2\times3\times3\times354=2×3×3×381=3\times3\times3\times381=3×3×3×3

therefore common factors between 18, 54, 81 is 3X3=9

xi) 12,45,75

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