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s344n2d4d5 [400]

2 years ago

11

If a+b+c=16 and a^2+b^2+c^2=90 then find the value a^3+b^3+c^3-3abc

1 answer:

grin007 [14]2 years ago

3 0

$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\\(16)^2=90+2(ab+bc+ca)\\256-90=2(ab+bc+ca)\\166/2=ab+bc+ca\\83=ab+bc+ca\\a^3-b^3-c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\(a+b+c)(a^2+b^2+c^2-(ab+bc+ca))\\16(90-(83))\\16(7)\\=112$

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V125BC [204]

Simple interest is given by:
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thus;
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aivan3 [116]

Hello and Good Morning/Afternoon

<u>Let's take this problem step-by-step</u>:

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⇒ <em>must set</em> the equations equal to each other

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<u>Let's put that into action</u>

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<u />

<u>Answer: (3,6), (-3,18)</u>

<u></u>

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#LearnwithBrainly

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