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Sagot :
Values of c and d make the equation true are c=6, d=2
Equations
- We must find the values of c and d that make the below equation be true
[tex]\sqrt[3]{162x^{c}y^{5} } = 3x^{2} y^{3} \sqrt[3]{6y^{d} }[/tex]
- cubing on both sides -
[tex](\sqrt[3]{162x^{c}y^{5} })^{3} = (3x^{2} y^{3} \sqrt[3]{6y^{d} })^{3}[/tex]
- The left side just simplifies the cubic root with the cube:
[tex]{162x^{c}y^{5} } = (3x^{2} y^{3} \sqrt[3]{6y^{d} })^{3}[/tex]
- On the right side, we'll simplify the cubic root where possible and power what's outside of the root:
[tex]{162x^{c}y^{5} } = 27x^{6} y^{3} ({6y^{d})[/tex]
- Simplifying
[tex]{x^{c}y^{5} } = x^{6} y^{3} ({y^{d})[/tex]
[tex]{x^{c}y^{5} } = x^{6} y^{3+d}[/tex]
- On equating,
c = 6
d = 2
To learn more about equations from the given link
https://brainly.com/question/14751707
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