find the cube root of the following number by prime factorisation 1)10648 2)15625
Answers
Answer:
Ncert solutions
Grade 8
Mathematics
Science
Chapters in NCERT Solutions - Mathematics , Class 8
Exercises in Cubes and Cube Roots
Question 1
Q1) Find the cube root of each of the following numbers by prime factorization method:
(i) 64
(ii) 512
(iii) 10648
(iv) 27000
(v) 15625
(vi) 13824
(vii) 110592
(viii) 46656
(ix) 175616
(x) 91125
Solution:
(i) 64
\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2}
3
64
=
3
2×2×2×2×2×2
\sqrt[3]{64}=\ 2\times2
3
64
= 2×2
= 4
(ii) 512
\sqrt[3]{512}=\sqrt[3]{2\times2\times2\times2\times2\times2\times2\times2\times2}
3
512
=
3
2×2×2×2×2×2×2×2×2
= 2 x 2 x 2 = 8
(iii) 10648
\sqrt[3]{10648}=\sqrt[3]{2\times2\times2\times11\times11\times11}
3
10648
=
3
2×2×2×11×11×11
= 2 x 11
= 22
(iv) 27000
\sqrt[3]{27000}=\sqrt[3]{2\times2\times2\times3\times3\times3\times5\times5\times5}
3
27000
=
3
2×2×2×3×3×3×5×5×5
= 2 x 3 x 5
= 30
(v) 15625
\sqrt[3]{15625}=\sqrt[3]{5\times5\times5\times5\times5\times5}
3
15625
=
3
5×5×5×5×5×5
= 5 x 5
= 25
(vi) 13824
\sqrt[3]{13824}=\sqrt[3]{2\times2\times2\times2\times2\times2\times2\times2\times2\times3\times3\times3}
3
13824
=
3
2×2×2×2×2×2×2×2×2×3×3×3
= 2 x 2 x 2 x 3
= 24
(vii) 110592
\sqrt[3]{110592}=\sqrt[3]{2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times3\times3\times3}
3
110592
=
3
2×2×2×2×2×2×2×2×2×2×2×2×3×3×3
= 2 x 2 x 2 x 2 x 3
= 48
(viii) 46656
\sqrt[3]{46645}=\sqrt[3]{2\times2\times2\times2\times2\times2\times3\times3\times3\times3\times3\times3}
3
46645
=
3
2×2×2×2×2×2×3×3×3×3×3×3
= 2 x 2 x 3 x 3
= 36
(ix) 175616
\sqrt[3]{175616}=\sqrt[3]{2\times2\times2\times2\times2\times2\times2\times2\times2\times7\times7\times7}
3
175616
=
3
2×2×2×2×2×2×2×2×2×7×7×7
= 2 x 2 x 2 x 7
= 56
(x) 91125
\sqrt[3]{91125}=\sqrt[3]{3\times3\times3\times3\times3\times3\times5\times5\times5}
3
91125
=
3
3×3×3×3×3×3×5×5×5
= 3 x 3 x 5 = 45
Answer:
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