
If the sum of two numbers is 11 and their cubes is 737, find the sum of their squares.
Answer
597.9k+ views
Hint: Consider two numbers a and b. Write the expression as per the condition in the question. Manipulate the expression and use the suitable identities that are given below to get the final answer.
\[1.\,{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab........\left( 1 \right)\]
$2.\,{{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab.........\left( 2 \right)$
$3.\,{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right).......\left( 3 \right)$
$4.\,{{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right).......\left( 4 \right)$
Complete step-by-step answer:
Let us consider two numbers as a and b.
The sum of these two numbers is 11. (Given in the problem)
Therefore, a+b=11…….(5)
The sum of cubes of these two numbers is 737. (Given in the problem)
Therefore, ${{a}^{3}}+{{b}^{3}}=737.......\left( 6 \right)$
Now, we have to find the sum of squares of these two numbers.
That is : ${{a}^{2}}+{{b}^{2}}=$?
Now, on taking cubes on both sides of equation 5, we get:
$\,\,\,\,\,{{\left( a+b \right)}^{3}}={{\left( 11 \right)}^{3}}$
Substituting the expression of ${{\left( a+b \right)}^{3}}$from the equation (3), we get:
${{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)=1331$
Substituting the value of ${{a}^{3}}+{{b}^{3}}=737$ from the equation (6) and the value of a+b = 11 from equation (5), we get:
737+3ab(11)=1331
On simplifying the above equation, we have:
33ab=1331-737
33ab=594
ab = 18……..(7)
Now, from the equation 2, we get:
${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$
On substituting the the value of a+b = 11 from equation (5) and the value of ab = 18 from equation (7) , we get:
${{a}^{2}}+{{b}^{2}}={{\left( 11 \right)}^{2}}-2\times 18$
On simplifying the above equation, we get:
${{a}^{2}}+{{b}^{2}}=121-36$
${{a}^{2}}+{{b}^{2}}=85$
Hence, the sum of squares of given two numbers is 85.
Note: In such types of problems, students are advised to remember the algebraic identities as it is useful in solving such types of problems. Students might get confused on taking identities. So be careful while using identity.
\[1.\,{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab........\left( 1 \right)\]
$2.\,{{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab.........\left( 2 \right)$
$3.\,{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right).......\left( 3 \right)$
$4.\,{{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right).......\left( 4 \right)$
Complete step-by-step answer:
Let us consider two numbers as a and b.
The sum of these two numbers is 11. (Given in the problem)
Therefore, a+b=11…….(5)
The sum of cubes of these two numbers is 737. (Given in the problem)
Therefore, ${{a}^{3}}+{{b}^{3}}=737.......\left( 6 \right)$
Now, we have to find the sum of squares of these two numbers.
That is : ${{a}^{2}}+{{b}^{2}}=$?
Now, on taking cubes on both sides of equation 5, we get:
$\,\,\,\,\,{{\left( a+b \right)}^{3}}={{\left( 11 \right)}^{3}}$
Substituting the expression of ${{\left( a+b \right)}^{3}}$from the equation (3), we get:
${{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)=1331$
Substituting the value of ${{a}^{3}}+{{b}^{3}}=737$ from the equation (6) and the value of a+b = 11 from equation (5), we get:
737+3ab(11)=1331
On simplifying the above equation, we have:
33ab=1331-737
33ab=594
ab = 18……..(7)
Now, from the equation 2, we get:
${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$
On substituting the the value of a+b = 11 from equation (5) and the value of ab = 18 from equation (7) , we get:
${{a}^{2}}+{{b}^{2}}={{\left( 11 \right)}^{2}}-2\times 18$
On simplifying the above equation, we get:
${{a}^{2}}+{{b}^{2}}=121-36$
${{a}^{2}}+{{b}^{2}}=85$
Hence, the sum of squares of given two numbers is 85.
Note: In such types of problems, students are advised to remember the algebraic identities as it is useful in solving such types of problems. Students might get confused on taking identities. So be careful while using identity.
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